题目

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

Many test cases will be given. 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.

Sample Input

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

2
3

题解

`i-begin+2` 就是前 `i` 个数在以第 `i` 个数结束的最短位置

代码

```/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！

*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <bitset>
using namespace std;

const int INF = 0x7FFFFFFF;
const int maxn = 100005;

int n,s;
int a[maxn];
long long sum[maxn];

bool Could(int n,int i) {
return sum[i] - sum[n - 1] < s;
}

int Division(int l,int r,int i) {
if(l == r) {
return l;
}
int mid = (l + r) / 2;
if(Could(mid,i))
return Division(l,mid,i);
else
return Division(mid + 1,r,i);
}

bool Do() {
if(!(cin >> n >> s))
return false;

int Min = INF;
sum[0] = 0;
for(int i = 1;i <= n;i++) {
cin >> a[i];
sum[i] = sum[i - 1] + a[i];
int begin = Division(1,i,i);
if(a[i] >= s)
Min = min(Min,1);
else if(begin != 1)
Min = min(Min,i - begin + 2);
}

if(Min == INF)
Min = 0;
cout << Min << endl;

return true;
}

int main() {
while(Do());
return 0;
}
```