POJ 3061 Subsequence【尺取法】

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10895		Accepted: 4507

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

恩，题意大致是，一序列，求最短的连续的和大于等于S的长度。

尺取法：1，初始化s=t=sum=0.

2，只要依然有sum<S，就不断将sum增加a(t)，并将t增加1.

3，如果2中无法满足sum>=S则终止。否则，更新ans=min(ans,t-s)。

4，将sum减去a(s)，s增加1然后回到2. ---《挑战程序设计》

#include <iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#define maxn 100010
using namespace std;
int a[maxn];
int main()
{
int t,n,s;
__int64 ss;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&s);
for(int i=1;i<=n;++i)
scanf("%d",&a[i]);
ss=a[1];
int i=0,j=1;
int ans=n+1;
int flag=1;
while(1)
{
while(ss<s)
{
j++;
if(j>n)
{
flag=0;
break;
}
ss+=a[j];
}
if(!flag)
break;
ans=min(ans,j-i);
ss=ss-a[++i];
}
if(ans==n+1)
printf("0\n");
else
printf("%d\n",ans);
}
return 0;
}