CodeForces 165C - Another Problem on Strings

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题意：给定一个长度为 n (1 <= n <= 10^6 ) 的01串，求有多少连续子序列中包含 k 个 1。

从前往后记录到第 i 个元素存在了多少个 1，然后从前往后找个数 >= k 的元素，再找 - k 与 - k + 1 之间与多少个元素加入结果即可。

#include<cstdio>#include<cstring>#include<cctype>#include<cstdlib>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<list>typedef long long ll;typedef unsigned long long llu;const int MAXN = 100 + 10;const int MAXT = 1000000 + 10;const int INF = 0x7f7f7f7f;const double pi = acos(-1.0);const double EPS = 1e-6;using namespace std;int n, k, num[MAXT];char s[MAXT];int main(){    scanf("%d%s", &k, s + 1);    n = strlen(s + 1);    num[0] = 0;    for(int i = 1; i <= n; ++i){        num[i] = num[i - 1];        if(s[i] == '1')  ++num[i];    }    num[n + 1] = num[n] + 1;    llu ans = 0;    for(int i = 1; i <= n; ++i)        if(num[i] >= k){            int c1 = lower_bound(num, num + i, num[i] - k) - num;            int c2 = upper_bound(num, num + i, num[i] - k) - num;            ans += llu(c2 - c1);        }    printf("%I64d\n", ans);    return 0;}

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