POJ 3061：Subsequence 查找连续的几个数，使得这几个数的和大于给定的S

Subsequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10172 Accepted: 4160

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

210 155 1 3 5 10 7 4 9 2 85 111 2 3 4 5

Sample Output

23

题意是给出一个序列，要求从这个序列中找出几个连续的数，使得这些数的和大于给定的S，求这个数的最小值。

和51nod上面的建设国家很像，都是建立一个队列，然后不断查找。

代码：

#include <iostream>#include <algorithm>#include <cmath>#include <vector>#include <string>#include <cstring>#pragma warning(disable:4996)using namespace std;long long n,s,a[1000005];int main(){//freopen("i.txt","r",stdin);//freopen("o.txt","w",stdout);int test,i,ans,start;long long sum;scanf("%d",&test);while(test--){scanf("%lld%lld",&n,&s);for(i=1;i<=n;i++){scanf("%lld",a+i);}ans=n+1;a[0]=0;sum=0;start=1;for(i=1;i<=n;i++){sum += a[i];while(sum>s){sum = sum - a[start];start++;}   if(sum+a[start-1]>s && i-(start-1)+1<ans)   {   ans=i-(start-1)+1;   }}if(ans == n+1){cout<<0<<endl;}else{cout<<ans<<endl;}}//system("pause");return 0;}