Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.   Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).   Output For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.   Sample Input 2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5   Sample Output Case 1: 14 1 4   Case 2: 7 1 6   題意：這題目是想要說明一串資料中，找出它的最大值，這個最大值得保證是某一個數到另一個數的和。例如：6 -1 5 4 -7，最大值便為14，其計算過程為（6-1+5+4）；0 6 -1 1 -6 7 -5，最大值便為7，其計算過程為（0+6-1+1-6+7）。這是比較簡單的DP題，但是這道題最關鍵的，在於換行的問題：具體的輸入輸出應該為： 2
5
6 -1 5 4 -7
Case 1:
14 1 4
7
0 6 -1 1 -6 7 -5   Case 2:
7 1 6   AC程式碼：
#include<stdio.h>
int dp[100005];
int main()
{
    int n,m,i,num;
    int sum,start,zuo,you;
    m=0;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&num);
        for(i=0;i<num;i++)
        {
            scanf("%d",&dp[i]);
        }
        start=0;
        zuo=0;
        you=0;
        sum=dp[0];
        for(i=1;i<num;i++)
        {
            if(dp[i-1]>=0)
            {
                dp[i]+=dp[i-1];
            }
            else
            {
                start=i;
            }
            if(dp[i]>sum)
            {
                sum=dp[i];
                zuo=start;
                you=i;
            }
        }
        if(m)
        {
            printf("\n");
        }
        printf("Case %d:\n",++m);
        printf("%d %d %d\n",sum,zuo+1,you+1);
    }
    return 0;
}