poj 3061 Subsequence -尺取法

Subsequence

Description

A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are given. Write a program to find the minimal length of the subsequence of consecutive elements of the sequence, the sum of which is greater than or equal to S.

Input

The first line is the number of test cases. For each test case the program has to read the numbers N and S, separated by an interval, from the first line. The numbers of the sequence are given in the second line of the test case, separated by intervals. The input will finish with the end of file.

Output

For each the case the program has to print the result on separate line of the output file.if no answer, print 0.

Sample Input

210 155 1 3 5 10 7 4 9 2 85 111 2 3 4 5

Sample Output

23

题意：

n个整数，求总和不小于S的连续子序列的长度的最小值，不存在输出0；

解题思路

尺取法

代码

#include <iostream>#include <cstdio>#include <map>#include <cmath>#include <string.h>#include <algorithm>#include <set>#include <sstream>#include <vector>#include <queue>#include <stack>using namespace std;const int maxn = 100000+10;const int INF = 0x3f3f3f3f;int main(){    int t;    scanf("%d",&t);    while(t > 0){        t--;        int n,S;        int a[maxn];        scanf("%d%d",&n,&S);        for(int i = 0;i < n; ++i){            scanf("%d",&a[i]);        }        int s = 0,t = 0;        int sum = 0;        int res = n+1;        while(1){            while(s < n && t < n && sum < S){                sum += a[t++];            }            if(sum < S) break;///走到最后了还是不满足            res = min(res,t-s);            ///左区间右移            sum -= a[s++];        }        if(res > n) res = 0;        printf("%d\n",res);    }    return 0;}