POJ-----3061---Subsequence---尺取法

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13982		Accepted: 5908

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

2
10 15
5 1 3 5 10 7 4 9 2 8
5 11
1 2 3 4 5

Sample Output

2
3

尺取法即是用来求一个序列最短的子串长度，且这个子串所有元素之和大于等于一个值

取左端点为序列最左侧的点，然后右端点依次向右取，当所有元素的值大于规定值时，更新最短子串长度，左端点向右移动一位，依次进行下去即可

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<queue>
#include<cmath>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long  LL;
const int maxn = 1e5+10;
const int mod = 1e9+7;
int s[maxn];
int main(){
int t, n, k, ans, sum;
cin >> t;
while(t--){
cin >> n >> k;
for(int i = 0; i < n; i++) cin >> s[i];
int i, j;
i = j = sum = 0;
ans = n+1;
while(1){
while(j < n && sum <= k) sum += s[j++];
if(sum < k) break;
ans = min(j-i, ans);
sum -= s[i++];
}
if(ans > n) ans = 0;
cout << ans;
}
return 0;
}