UVa 2678 Subsequence / 二分

求长度最短的连续序列 它的和大于等于s 输出长度

枚举起点和终点会超时

求出前缀和 都是正整数 所以前缀和是递增的

如果对于前缀和 sum[i]要使得长度最小 那么应该找出最大的j 使得 sum[i]-sum[j]>=s   sum[j] <= sum[i]-s 即二分sum[i]-s 找到最大的sum[j]

 

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 100010;int a[maxn];int sum[maxn];int main(){int n, s;while(scanf("%d %d", &n, &s) ==2 ){for(int i = 1; i <= n; i++){scanf("%d", &a[i]);sum[i] = sum[i-1] + a[i];}int ans = n+1;for(int i = 1; i <= n; i++){int l = 0;int r = i;int k = sum[i] - s;if(k < 0)continue;while(l <= r){int m = (l + r) >> 1;if(sum[m] <= k)l = m + 1;elser = m - 1;}ans = min(ans, i-l+1);}if(ans == n+1)ans = 0;printf("%d\n", ans);}return 0;}