lightoj1080

maksyuki 发表于 oj 分类，标签: 数据结构-线段树

27 8月 2015

1080 - Binary Simulation

'I i j' which means invert the bit from i to j (inclusive)Given a binary number, we are about to do some operations on the number. Two types of operations can be here.

'Q i' answer whether the i^th bit is 0 or 1

The MSB (most significant bit) is the first bit (i.e. i=1). The binary number can contain leading zeroes.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing a binary integer having length n (1 ≤ n ≤ 10⁵). The next line will contain an integer q (1 ≤ q ≤ 50000) denoting the number of queries. Each query will be either in the form 'I i j' where i, j are integers and 1 ≤ i ≤ j ≤ n. Or the query will be in the form 'Q i' where i is an integer and 1 ≤ i ≤ n.

Output

For each case, print the case number in a single line. Then for each query 'Q i' you have to print 1 or 0 depending on the i^th bit.

Sample Input

Output for Sample Input

200110011006I 1 10

I 2 7

Q 2

Q 1

Q 7

Q 5

1011110111

I 1 10

I 2 7

Q 2

Q 1

Q 7

Q 5

Case 1:011

Case 2:

Note

Dataset is huge, use faster i/o methods.

题目类型：线段树

算法分析：需要使用lazy标记来进行区间修改，lazy布尔数组当为false时表示当前区间没有翻转，而true表示当前区间元素需要翻转。由于本题要进行单点查询，所以可以在查询到叶子节点后再判断lazy，且不需要PushUp

#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>

#define lson rt << 1, l, m
#define rson rt << 1 | 1, m + 1, r

using namespace std;

const int INF = 0x7FFFFFFF;
const int MOD = 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 100000 + 66;

bool lazy[maxn<<2];
char sum[maxn];

void PushDown (int rt)
{
    if (lazy[rt])
    {
        lazy[rt<<1] = !lazy[rt<<1];
        lazy[rt<<1|1] = !lazy[rt<<1|1];
        lazy[rt] = !lazy[rt];
    }
}


void UpDate (int rt, int l, int r, int L, int R)
{
    if (L <= l && r <= R)
    {
        lazy[rt] = !lazy[rt];
        return ;
    }

    int m = (l + r) >> 1;
    PushDown (rt);

    if (L <= m)
        UpDate (lson, L, R);

    if (R > m)
        UpDate (rson, L, R);

}

int Query (int rt, int l, int r, int val)
{
   if (l == r)
   {
       if (lazy[rt])
       {
           sum[val] = !(sum[val] - '0') + '0';
           lazy[rt] = !lazy[rt];
       }

       return sum[val] - '0';
   }

    int m = (l + r) >> 1;
    PushDown (rt);

   if (val <= m)
        return Query (lson, val);

   else
        return Query (rson, val);
}


int main()
{
//    freopen ("aaa.txt", "r", stdin);

    int t, flag = 1;
    scanf ("%d", &t);

    while (t--)
    {
        printf ("Case %d:\n", flag++);

        memset (lazy, false, sizeof (lazy));
        scanf ("%s", sum + 1);

        int q, len = strlen (sum + 1);

        scanf ("%d", &q);

        int i;
        char cmd[6];
        for (i = 0; i < q; i++)
        {
            scanf ("%s", cmd);
            if (cmd[0] == 'I')
            {
                int val_a, val_b;
                scanf ("%d%d", &val_a, &val_b);
                UpDate (1, 1, len, val_a, val_b);
            }

            else
            {
                int val;
                scanf ("%d", &val);
                printf ("%d\n", Query (1, 1, len, val));
            }
        }
    }

return 0;
}

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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>

#define lson rt << 1, l, m

#define rson rt << 1 | 1, m + 1, r

using namespace std;

const int INF = 0x7FFFFFFF;

const int MOD = 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 100000 + 66;

bool lazy[maxn<<2];

char sum[maxn];

void PushDown (int rt)

{

if (lazy[rt])

{

lazy[rt<<1] = !lazy[rt<<1];

lazy[rt<<1|1] = !lazy[rt<<1|1];

lazy[rt] = !lazy[rt];

}

void UpDate (int rt, int l, int r, int L, int R)

{

if (L <= l && r <= R)

{

lazy[rt] = !lazy[rt];

return ;

}

int m = (l + r) >> 1;

PushDown (rt);

if (L <= m)

UpDate (lson, L, R);

if (R > m)

UpDate (rson, L, R);

}

int Query (int rt, int l, int r, int val)

{

if (l == r)

{

if (lazy[rt])

{

sum[val] = !(sum[val] - '0') + '0';

lazy[rt] = !lazy[rt];

}

return sum[val] - '0';

}

int m = (l + r) >> 1;

PushDown (rt);

if (val <= m)

return Query (lson, val);

else

return Query (rson, val);

}

int main()

{

// freopen ("aaa.txt", "r", stdin);

int t, flag = 1;

scanf ("%d", &t);

while (t--)

{

printf ("Case %d:\n", flag++);

memset (lazy, false, sizeof (lazy));

scanf ("%s", sum + 1);

int q, len = strlen (sum + 1);

scanf ("%d", &q);

int i;

char cmd[6];

for (i = 0; i < q; i++)

{

scanf ("%s", cmd);

if (cmd[0] == 'I')

{

int val_a, val_b;

scanf ("%d%d", &val_a, &val_b);

UpDate (1, 1, len, val_a, val_b);

}

else

{

int val;

scanf ("%d", &val);

printf ("%d\n", Query (1, 1, len, val));

}

return 0;

}

lightoj1078

lightoj1083

超客的水墨烟雨

lightoj1080

Input

Output

Sample Input

Output for Sample Input

Note

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