CodeForces 165C - Another Problem on Strings(思路)
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; }