2 | 12月 | 2017 | 超客的水墨烟雨

uva10340

maksyuki 发表于 oj 分类，标签: 字符串-简单字符处理

02 12月 2017

You have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way. Because of pending patent issues we will not discuss in detail how the strings are generated and inserted into the original message. To validate your method, however, it is necessary to write a program that checks if the message is really encoded in the final string.

Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.

Input

The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace. Input is terminated by EOF.

Output

For each test case output, if s is a subsequence of t.

Sample Input

sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter

Sample Output

Yes

题目类型：简单字符处理

算法分析：直接顺序查找两个字符串是否有对应字符即可

/**************************************************
filename       :o.cpp
author         :maksyuki
created time   :2017/12/2 20:53:49
last modified  :2017/12/2 20:56:46
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e5 + 6666;

char s1[maxn], s2[maxn];

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif
    
    while(scanf(" %s %s", s1, s2) != EOF) {
        int len1 = strlen(s1), len2 = strlen(s2);
        if(len1 > len2) puts("No");
        else {
            int i, j;
            for(i = 0, j = 0; i < len1 && j < len2; ) {
                if(s1[i] == s2[j]) {
                    i++, j++;
                }
                else j++;
            }
            if(i == len1) puts("Yes");
            else puts("No");
        }
    }
    return 0;
}

/**************************************************

filename :o.cpp

author :maksyuki

created time :2017/12/2 20:53:49

last modified :2017/12/2 20:56:46

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e5 + 6666;

char s1[maxn], s2[maxn];

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

while(scanf(" %s %s", s1, s2) != EOF) {

int len1 = strlen(s1), len2 = strlen(s2);

if(len1 > len2) puts("No");

else {

int i, j;

for(i = 0, j = 0; i < len1 && j < len2; ) {

if(s1[i] == s2[j]) {

i++, j++;

}

else j++;

}

if(i == len1) puts("Yes");

else puts("No");

}

return 0;

}

uva227

maksyuki 发表于 oj 分类，标签: 基础算法-模拟

02 12月 2017

A children’s puzzle that was popular 30 years ago consisted of a 5×5 frame which contained 24 small squares of equal size. A unique letter of the alphabet was printed on each small square. Since there were only 24 squares within the frame, the frame also contained an empty position which was the same size as a small square. A square could be moved into that empty position if it were immediately to the right, to the left, above, or below the empty position. The object of the puzzle was to slide squares into the empty position so that the frame displayed the letters in alphabetical order.

The illustration below represents a puzzle in its original configuration and in its configuration after the following sequence of 6 moves:

1) The square above the empty position moves.

2) The square to the right of the empty position moves.

3) The square to the right of the empty position moves.

4) The square below the empty position moves.

5) The square below the empty position moves.

6) The square to the left of the empty position moves.

Write a program to display resulting frames given their initial configurations and sequences of moves.

Input

Input for your program consists of several puzzles. Each is described by its initial configuration and the sequence of moves on the puzzle. The first 5 lines of each puzzle description are the starting configuration. Subsequent lines give the sequence of moves.

The first line of the frame display corresponds to the top line of squares in the puzzle. The other lines follow in order. The empty position in a frame is indicated by a blank. Each display line contains exactly 5 characters, beginning with the character on the leftmost square (or a blank if the leftmost square is actually the empty frame position). The display lines will correspond to a legitimate puzzle.

The sequence of moves is represented by a sequence of As, Bs, Rs, and Ls to denote which square moves into the empty position. A denotes that the square above the empty position moves; B denotes that the square below the empty position moves; L denotes that the square to the left of the empty position moves; R denotes that the square to the right of the empty position moves. It is possible that there is an illegal move, even when it is represented by one of the 4 move characters. If an illegal move occurs, the puzzle is considered to have no final configuration. This sequence of moves may be spread over several lines, but it always ends in the digit 0. The end of data is denoted by the character Z.

Output

Output for each puzzle begins with an appropriately labeled number (Puzzle #1, Puzzle #2, etc.). If the puzzle has no final configuration, then a message to that effect should follow. Otherwise that final configuration should be displayed.

Format each line for a final configuration so that there is a single blank character between two adjacent letters. Treat the empty square the same as a letter. For example, if the blank is an interior position, then it will appear as a sequence of 3 blanks — one to separate it from the square to the left, one for the empty position itself, and one to separate it from the square to the right.

Separate output from different puzzle records by one blank line.

Note: The first record of the sample input corresponds to the puzzle illustrated above.

Sample Input

TRGSJ
XDOKI
M VLN
WPABE
UQHCF
ARRBBL0
ABCDE
FGHIJ
KLMNO
PQRS
TUVWX
AAA
LLLL0
ABCDE
FGHIJ
KLMNO
PQRS
TUVWX
AAAAABBRRRLL0
Z

Sample Output

Puzzle #1:
T R G S J
X O K L I
M D V B N
W P A E
U Q H C F
Puzzle #2:
A B C D
F G H I E
K L M N J
P Q R S O
T U V W X
Puzzle #3:
This puzzle has no final configuration.

题目类型：简单模拟

算法分析：按照要求顺序模拟移动，注意输入的方法

/**************************************************
filename       :k.cpp
author         :maksyuki
created time   :2017/12/2 10:33:51
last modified  :2017/12/2 11:15:31
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 66;

char g[maxn][maxn];
const char v[]= {'A', 'B', 'L', 'R'};
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};

bool check(int x, int y) {
    if(x >= 0 && x <= 4 && y >= 0 && y <= 4) return true;
    return false;
}

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif

    int id = 1;
    while(fgets(g[0], maxn, stdin)) {
        if(g[0][0] == 'Z') break;

        for(int i = 1; i < 5; i++) fgets(g[i], maxn, stdin);
        int c;
        bool is_valid = true;
        int px = 0, py = 0;
        for(int i = 0; i < 5; i++)
            for(int j = 0; j < 5; j++)
                if(g[i][j] == ' ') px = i, py = j;
        
        while((c = fgetc(stdin)) != '0') {
            for(int i = 0; i < 4; i++) {
                if(c == v[i]) {
                    int tmpx = px + dx[i], tmpy = py + dy[i];
                    if(check(tmpx, tmpy)) {
                        swap(g[tmpx][tmpy], g[px][py]);
                        px = tmpx, py = tmpy;
                    }
                    else is_valid = false;
                }
            }
        }

        if(id > 1) puts("");
        printf("Puzzle #%d:\n", id++);

        if(!is_valid) puts("This puzzle has no final configuration.");
        
        else {
            for(int i = 0; i < 5; i++) {
                for(int j = 0; j < 5; j++) {
                    if(!j) printf("%c", g[i][j]);
                    else printf(" %c", g[i][j]);
                }
                puts("");
            }
        }

        fgets(g[0], maxn, stdin);
    }
    return 0;
}

100

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119

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121

/**************************************************

filename :k.cpp

author :maksyuki

created time :2017/12/2 10:33:51

last modified :2017/12/2 11:15:31

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 66;

char g[maxn][maxn];

const char v[]= {'A', 'B', 'L', 'R'};

const int dx[] = {-1, 1, 0, 0};

const int dy[] = {0, 0, -1, 1};

bool check(int x, int y) {

if(x >= 0 && x <= 4 && y >= 0 && y <= 4) return true;

return false;

}

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

int id = 1;

while(fgets(g[0], maxn, stdin)) {

if(g[0][0] == 'Z') break;

for(int i = 1; i < 5; i++) fgets(g[i], maxn, stdin);

int c;

bool is_valid = true;

int px = 0, py = 0;

for(int i = 0; i < 5; i++)

for(int j = 0; j < 5; j++)

if(g[i][j] == ' ') px = i, py = j;

while((c = fgetc(stdin)) != '0') {

for(int i = 0; i < 4; i++) {

if(c == v[i]) {

int tmpx = px + dx[i], tmpy = py + dy[i];

if(check(tmpx, tmpy)) {

swap(g[tmpx][tmpy], g[px][py]);

px = tmpx, py = tmpy;

}

else is_valid = false;

}

if(id > 1) puts("");

printf("Puzzle #%d:\n", id++);

if(!is_valid) puts("This puzzle has no final configuration.");

else {

for(int i = 0; i < 5; i++) {

for(int j = 0; j < 5; j++) {

if(!j) printf("%c", g[i][j]);

else printf(" %c", g[i][j]);

}

puts("");

}

fgets(g[0], maxn, stdin);

}

return 0;

}

uva1586

maksyuki 发表于 oj 分类，标签: 字符串-简单字符处理

02 12月 2017

An organic compound is any member of a large class of chemical compounds whose molecules contain carbon. The molar mass of an organic compound is the mass of one mole of the organic compound. The molar mass of an organic compound can be computed from the standard atomic weights of the elements.

When an organic compound is given as a molecular formula, Dr. CHON wants to find its molar mass. A molecular formula, such as C3H4O3, identifies each constituent element by its chemical symbol and indicates the number of atoms of each element found in each discrete molecule of that compound. If a molecule contains more than one atom of a particular element, this quantity is indicated using a subscript after the chemical symbol.

In this problem, we assume that the molecular formula is represented by only four elements, ‘C’ (Carbon), ‘H’ (Hydrogen), ‘O’ (Oxygen), and ‘N’ (Nitrogen) without parentheses.

The following table shows that the standard atomic weights for ‘C’, ‘H’, ‘O’, and ‘N’.

Atomic Name Carbon Hydrogen Oxygen Nitrogen

Standard Atomic Weight 12.01 g/mol 1.008 g/mol 16.00 g/mol 14.01 g/mol

For example, the molar mass of a molecular formula C6H5OH is 94.108 g/mol which is computed by 6 × (12.01 g/mol) + 6 × (1.008 g/mol) + 1 × (16.00 g/mol).

Given a molecular formula, write a program to compute the molar mass of the formula.

Input

Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case is given in a single line, which contains a molecular formula as a string. The chemical symbol is given by a capital letter and the length of the string is greater than 0 and less than 80. The quantity number n which is represented after the chemical symbol would be omitted when the number is 1 (2 ≤ n ≤ 99).

Output

Your program is to write to standard output. Print exactly one line for each test case. The line should contain the molar mass of the given molecular formula.

Sample Input

4
C
C6H5OH
NH2CH2COOH
C12H22O11

Sample Output

12.010

94.108

75.070

342.296

题目类型：简单字符处理

算法分析：直接顺序累加计算即可

/**************************************************
filename       :h.cpp
author         :maksyuki
created time   :2017/12/2 8:50:06
last modified  :2017/12/2 9:12:44
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e5 + 6666;

const char c[] = {'*', 'C', 'H', 'O', 'N'};
const double v[] = {-6.00, 12.01, 1.008, 16.00, 14.01};
int cnt[6];
char s[maxn];

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif
    
    int t;
    scanf("%d", &t);
    while(t--) {
        scanf("%s", s);
        
        int tmp = 0, last = 1;
        memset(cnt, 0, sizeof(cnt));
        for(int i = 0; s[i]; ) {
            for(int j = 1; j <= 4; j++) {
                if(s[i] && s[i] == c[j]) {
                    last = j;
                    i++;
                    while(s[i] && isdigit(s[i])) {
                        tmp = tmp * 10 + (s[i] - '0');
                        i++;
                    }
                    if(!tmp) tmp = 1;
                    cnt[j] += tmp;
                    tmp = 0;
                }
            }
        }
        if(tmp) cnt[last] += tmp;
        
        double res = 0;
        for(int i = 1; i <= 4; i++) res += v[i] * cnt[i];
        printf("%.3lf\n", res);
    }
    return 0;
}

100

101

102

/**************************************************

filename :h.cpp

author :maksyuki

created time :2017/12/2 8:50:06

last modified :2017/12/2 9:12:44

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e5 + 6666;

const char c[] = {'*', 'C', 'H', 'O', 'N'};

const double v[] = {-6.00, 12.01, 1.008, 16.00, 14.01};

int cnt[6];

char s[maxn];

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

int t;

scanf("%d", &t);

while(t--) {

scanf("%s", s);

int tmp = 0, last = 1;

memset(cnt, 0, sizeof(cnt));

for(int i = 0; s[i]; ) {

for(int j = 1; j <= 4; j++) {

if(s[i] && s[i] == c[j]) {

last = j;

i++;

while(s[i] && isdigit(s[i])) {

tmp = tmp * 10 + (s[i] - '0');

i++;

}

if(!tmp) tmp = 1;

cnt[j] += tmp;

tmp = 0;

}

if(tmp) cnt[last] += tmp;

double res = 0;

for(int i = 1; i <= 4; i++) res += v[i] * cnt[i];

printf("%.3lf\n", res);

}

return 0;

}

uva1585

maksyuki 发表于 oj 分类，标签: 字符串-简单字符处理

02 12月 2017

There is an objective test result such as “OOXXOXXOOO”. An ‘O’ means a correct answer of a problem and an ‘X’ means a wrong answer. The score of each problem of this test is calculated by itself and its just previous consecutive ‘O’s only when the answer is correct. For example, the score of the 10th problem is 3 that is obtained by itself and its two previous consecutive ‘O’s.

Therefore, the score of “OOXXOXXOOO” is 10 which is calculated by “1+2+0+0+1+0+0+1+2+3”.

You are to write a program calculating the scores of test results.

Input

Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case starts with a line containing a string composed by ‘O’ and ‘X’ and the length of the string is more than 0 and less than 80. There is no spaces between ‘O’ and ‘X’.

Output

Your program is to write to standard output. Print exactly one line for each test case. The line is to contain the score of the test case.

Sample Input

5
OOXXOXXOOO
OOXXOOXXOO
OXOXOXOXOXOXOX
OOOOOOOOOO
OOOOXOOOOXOOOOX

Sample Output

题目类型：简单字符处理

算法分析：按照顺序依次累加即可

/**************************************************
filename       :g.cpp
author         :maksyuki
created time   :2017/12/2 8:26:38
last modified  :2017/12/2 8:33:42
file location  :C:\Users\abcd\Desktop\TheEternalPoet
***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("in", "r", stdin)
#define CFO freopen ("out", "w", stdout)
#define CPPFF ifstream cin ("in")
#define CPPFO ofstream cout ("out")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define	DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e5 + 6666;

char s[maxn];

int main()
{
#ifdef LOCAL
    CFF;
    //CFO;
#endif
    
    int t;
    scanf("%d", &t);

    while(t--) {
       scanf("%s", s); 

       int res = 0, tmp = 0;
       for(int i = 0; s[i]; ) {
           if(s[i] == 'O') {
                int cnt = 1;
                do {
                    tmp += cnt;
                    cnt++, i++; 
                }while(s[i] && s[i] == 'O');
           }
           else {
               res += tmp;
               tmp = 0;
               i++;
           }
       }

       if(tmp) res += tmp;
       printf("%d\n", res);
    }
    return 0;
}

/**************************************************

filename :g.cpp

author :maksyuki

created time :2017/12/2 8:26:38

last modified :2017/12/2 8:33:42

file location :C:\Users\abcd\Desktop\TheEternalPoet

***************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("in", "r", stdin)

#define CFO freopen ("out", "w", stdout)

#define CPPFF ifstream cin ("in")

#define CPPFO ofstream cout ("out")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC)

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e5 + 6666;

char s[maxn];

int main()

{

#ifdef LOCAL

CFF;

//CFO;

#endif

int t;

scanf("%d", &t);

while(t--) {

scanf("%s", s);

int res = 0, tmp = 0;

for(int i = 0; s[i]; ) {

if(s[i] == 'O') {

int cnt = 1;

do {

tmp += cnt;

cnt++, i++;

}while(s[i] && s[i] == 'O');

}

else {

res += tmp;

tmp = 0;

i++;

}

if(tmp) res += tmp;

printf("%d\n", res);

}

return 0;

}

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