poj Subsequence 3061 (高效&DP)

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10880		Accepted: 4497

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

//题意：

给你两个数n,m，n表示有n个数，m表示目标分数，问最短的连续数列的和>=m,求出这个数。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#define INF 0x3f3f3f3f
#define ll long long
#define N 100010
using namespace std;
ll sum
;
int main()
{
int t,n,m,a;
int i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
{
scanf("%d",&a);
sum[i]=sum[i-1]+a;
}
int mm=n+1;
for(i=j=1;j<=n;j++)
{
if(sum[j]-sum[i-1]<m)
continue;
while(sum[j]-sum[i]>=m)
i++;
mm=min(mm,j-i+1);
}
if(mm==n+1)
printf("0\n");
else
printf("%d\n",mm);
}
return 0;
}