poj 3061

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给定长度为n的数列整数a0,a1,。。。an-1以及整数S，求出总和不小于S的连续子序列长度的最小值。如果解不存在，则输出零。

刷挑战。。。

尺取和二分，果然还是尺取块。。。

#include<iostream>#include<algorithm>#include<string>#include<queue>#include<cmath>#include<vector>#include<stdlib.h>#include<iomanip>#include<list>#include<stack>#include<memory.h>#include<ctype.h>using namespace std;typedef long long ll;const int maxn = 1e5 + 5;int num[maxn];int n, m;void solve() {//尺取int ans = n + 1;int start = 0, t = 0, sum = 0;for (;;) {while (t < n&&sum < m) {sum += num[t++];}if (sum < m)break;ans = min(ans, t - start);sum -= num[start++];}if (ans > n)ans = 0;printf("%d\n", ans);}int main() {int t;scanf("%d", &t);while (t--) {scanf("%d%d", &n, &m);for (int i = 0; i < n; i++) {scanf("%d", &num[i]);}solve();}}/*int main() {//二分int t;scanf("%d", &t);while (t--) {scanf("%d%d", &n, &m);int x;for (int i = 0; i <n; i++) {scanf("%d", &x);num[i + 1] = num[i] + x;}int ans = n;if (num[n] < m)ans = 0;else {for (int i = 0; num[i] + m <= num[n];i++) {x = lower_bound(num + i, num + n, num[i] + m)-num;//ForwardIter lower_bound(ForwardIter first, ForwardIter last,const _Tp& val)//算法返回一个非递减序列[first, last)中的第一个大于等于值val的位置。ans = min(ans, x-i);}}printf("%d\n", ans);}}*/

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