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
题意：
求出总和不小于S的连续子序列的长度的最小值。
题解：
二分查找的话，前缀和是满足单调性的，计算从每一个数开始总和刚好大于s的长度。具体实现就是
二分搜索s[i]+s是否存在于前缀和数组中，就是查找以i+1开头的总和刚好大于s的最短长度。
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxn=1e5+5;
int sum[maxn];
int main()
{
int t;
cin>>t;
while(t--)
{
int n,S;
cin>>n>>S;
for(int i=0;i<n;i++)
{
int x;
cin>>x;
sum[i+1]=sum[i]+x;
}
if(sum[n]<S)
{
puts("0");
continue;
}
int res=n;
for(int s=0;sum[s]+S<=sum[n];s++)
{
int t=lower_bound(sum+s,sum+n,sum[s]+S)-sum;
res=min(res,t-s);
}
cout<<res<<endl;
}
return 0;
}
解法二：尺取法
反复地推进区间的开头和末尾，来求取条件的最小区间的方法被称为尺取法。
因为很懒，所以直接贴图片了。
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1e5+5;
int a[maxn];
int main()
{
int t;
cin>>t;
while(t--)
{
int n,k;
cin>>n>>k;
int s=0,t=0,sum=0;
for(int i=0;i<n;i++)
cin>>a[i];
int res=n+1;
for(;;)
{
while(t<n&&sum<k)
sum+=a[t++];
if(sum<k)
break;
res=min(res,t-s);
sum-=a[s++];
}
if(res>n)
res=0;
cout<<res<<endl;
}
return 0;
}