POJ3061 Subsequence

题目大意：给定长度为n的整列整数a[0],a[1],……a[n-1]，以及整数S，求出总和不小于S的连续子序列的长度的最小值。

PS：用二分或者直尺法。省赛热身。

#include <cstdio>#include <cstdlib>#include <vector>#include <algorithm>#include <cstring>using namespace std;const int INF = 0x3fffff;vector<int> v;int n, S;int work() {int ans = INF;int l=0, r = 0;int sum = 0;for(int i = 0; i < n && sum < S; i++) {sum += v[i];r = i;}if(sum < S) return 0;while(r < n && l <= r) {    if(sum<S && r < n) {sum += v[++r];}if(sum >= S && r<n) {ans = min(ans, r-l+1);sum = sum - v[l++];}}return ans;}int main(){int T;//freopen("in", "r", stdin);scanf("%d", &T);int t;while(T--) {scanf("%d%d", &n, &S);v.clear();for(int i = 0; i < n; i++) {scanf("%d", &t);v.push_back(t);}int res = work();printf("%d\n", res);}    return 0;}