A.Kyoya and Photobooks
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
Input
The first line of input will be a single string s (1 ≤ |s| ≤ 20). String s consists only of lowercase English letters.
Output
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Sample test(s)
input
a
output
51
input
hi
output
76
Note
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
题目类型：简单模拟
算法分析：直接推出规律计算即可
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#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; const int INF = 0x7FFFFFFF; const int MOD = 1000000000 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 100000 + 66; int main() { // ifstream cin ("aaa.txt"); string s; cin >> s; int len = s.size (); len++; cout << 26 + (len - 1) * 25 << endl; return 0; } |
B.Ohana Cleans Up
Ohana Matsumae is trying to clean a room, which is divided up into an n by n grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column.
Return the maximum number of rows that she can make completely clean.
Input
The first line of input will be a single integer n (1 ≤ n ≤ 100).
The next n lines will describe the state of the room. The i-th line will contain a binary string with n characters denoting the state of the i-th row of the room. The j-th character on this line is '1' if the j-th square in the i-th row is clean, and '0' if it is dirty.
Output
The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.
Sample test(s)
input
4
0101
1000
1111
0101
output
2
input
3
111
111
111
output
3
Note
In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean.
In the second sample, everything is already clean, so Ohana doesn't need to do anything.
题目类型：简单模拟
算法分析：首先计算一下原来有多少行是都是1，记为maxval。根据题目的要求可知，只有两个完全相同的行才可以通过列变换来同步变成都是1的形式。所以只要找到行相同的最大值，然后与原来有多少1比较即可
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#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; const int INF = 0x7FFFFFFF; const int MOD = 1000000000 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 100 + 66; char a[maxn][maxn]; int main() { // ifstream cin ("aaa.txt"); int n; cin >> n; for (int i = 0; i < n; i++) cin >> a[i]; int maxval = 0; for (int i = 0; i < n; i++) { int j; for (j = 0; j < n; j++) if (a[i][j] == '0') break; if (j == n) maxval++; } int cnt = 0; for (int i = 0; i < n; i++) { int temp = 0; for (int j = i + 1; j < n; j++) { if (!strcmp (a[i], a[j])) temp++; } cnt = max (cnt, temp+1); } cout << max (maxval, cnt) << endl; return 0; } |
C.Kyoya and Colored Balls
Kyoya Ootori has a bag with n colored balls that are colored with k different colors. The colors are labeled from 1 tok. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color i before drawing the last ball of color i + 1 for all i from 1 to k - 1. Now he wonders how many different ways this can happen.
Input
The first line of input will have one integer k (1 ≤ k ≤ 1000) the number of colors.
Then, k lines will follow. The i-th line will contain ci, the number of balls of the i-th color (1 ≤ ci ≤ 1000).
The total number of balls doesn't exceed 1000.
Output
A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1 000 000 007.
Sample test(s)
input
3
2
2
1
output
3
input
4
1
2
3
4
output
1680
Note
In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are:
1 2 1 2 3
1 1 2 2 3
2 1 1 2 3
题目类型：组合+DP
算法分析：使用res表示取前i种颜色时的方案数，则i = 1时，res = 1;然后递推方程为res = res * Com (sum + cnt[i] – 1, cnt[i] – 1) % MOD，其中sum表示小于i的所有的颜色的球的个数(前缀和)，注意使用组合数取模！！！！！！！
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#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; const int INF = 0x7FFFFFFF; const int MOD = 1000000000 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1000 + 66; int cnt[maxn]; long long C[maxn][maxn]; void GetComMod (long long val, long long mod) { memset (C, 0, sizeof (C)); for (long long i = 0; i <= val; i++) { for (long long j = 0; j <= i; j++) { if (!j || i == j) C[i][j] = 1 % mod; else C[i][j] = (C[i-1][j-1] % mod + C[i-1][j] % mod) % mod; } } C[0][0] = 0 % mod; } int main() { GetComMod (1000 + 2, MOD); // ifstream cin ("aaa.txt"); int k; cin >> k; for (int i =1; i <= k; i++) cin >> cnt[i]; long long res = 1, tempsum = cnt[1]; for (int i = 2; i <= k; i++) { res = (res % MOD * C[tempsum+cnt[i]-1][cnt[i]-1]) % MOD; tempsum += cnt[i]; } cout << res << endl; return 0; } |