YTU 1020: I think it

1020: I think it

时间限制: 1 Sec  内存限制: 32 MB

提交: 501  解决: 63

题目描述

Xiao Ming is only seven years old, Now I give him some numbers, and ask him what is the second largest sum if he can choose a part of them. For example, if I give him 1 、 2 、 3 ， then he should tell me 5 as 6 is the largest and 5 is the second. I think
it is too hard for him, isn ’ t it?

输入

Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <=10) which is the number of test cases. And it will be followed by T consecutive test cases.

Each test case starts with a line containing an integer N (1<N<10) , the number I give Xiao Ming . The second line contains N Integer numbers ai (-10<ai<10),

输出

For each test case, output the answer.

样例输入

2
3
1 2 3
4
0 1 2 3

样例输出

5
5

#include<stdio.h>
int jisuan(int a[],int n)
{
    int s,ma,i,j,k,b;
    s=ma=a[0];
    for(i=0; i<n; i++)
        for(j=i; j<n; j++)
        {
            b=0;
            for(k=i; k<=j; k++)b+=a[k];
            if(b>ma)ma=b;
        }
    for(i=0; i<n; i++)
        for(j=i; j<n; j++)
        {
            b=0;
            for(k=i; k<=j; k++)b+=a[k];
            if(b>s&&b<ma)s=b;
        }
    return s;
}
int main()
{
    int N,n,a[15],i,t,j;
    scanf("%d",&N);
    while(N--)
    {
        scanf("%d",&n);
        for(i=0; i<n; i++)scanf("%d",&a[i]);
        for(i=0; i<n-1; i++)
            for(j=0; j<n-i-1; j++)
                if(a[j]>a[j+1])
                {
                    t=a[j];
                    a[j]=a[j+1];
                    a[j+1]=t;
                }
        printf("%d\n",jisuan(a,n));
    }
    return 0;
}

总是望着曾经的空间发呆，那些说好不分开的朋友不在了，转身，陌路。 熟悉的，安静了， 安静的，离开了， 离开的，陌生了， 陌生的，消失了， 消失的，陌路了。