Minimal Subarray Length

You are given an integer sequence of length N and another value X. You have to find a contiguous
subsequence of the given sequence such that the sum is greater or equal to X. And you have to find
that segment with minimal length.
Input
First line of the input file contains T the number of test cases. Each test case starts with a line
containing 2 integers N (1 ≤ N ≤ 500000) and X (−109 ≤ X ≤ 109
). Next line contains N integers
denoting the elements of the sequence. These integers will be between −109
to 109
inclusive.
Output
For each test case output the minimum length of the sub array whose sum is greater or equal to X. If
there is no such array, output ‘-1’.
Sample Input
3
5 4
1 2 1 2 1
6 -2
-5 -6 -7 -8 -9 -10
5 3
-1 1 1 1 -1
Sample Output
3
-1

3

题意：求关于数组的子串，其中最小长度满足条件的长度

（个人能力有限，目前只知道这种投机取巧的方法）

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int num[500005];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n,x;        scanf("%d%d",&n,&x);        for(int i=0;i<n;++i)        {            scanf("%d",&num[i]);        }        if(x<=0)        {            int c=0;            for(int i=0;i<n;++i)            {                if(num[i]>=x){c=1;break;}            }            if(c)printf("1\n");            else printf("-1\n");            continue;        }        int ans=1000000000,start=0;        long long  sum=0;        for(int i=0;i<n;++i)        {            sum=sum+num[i];            if(sum<=0)            {                sum=0;                start=i+1;            }            if(sum>=x)            {           int tmp=0;                for(int j=i;j>=0;--j)                {                    if(tmp+num[j]>=x)                    {                        start=j;                        break;                    }                    tmp+=num[j];                }                ans=min(ans,i-start+1);            }        }        if(ans>500001)printf("-1\n");        else            printf("%d\n",ans);    }    return 0;}