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

 

Sample Output

解题心得：

1、这个题要求的是一个在序列中的一串连续的子序列，要求子序列的和最大，并且要求记录子序列的起始位置和终止位置。

2、可以将前面的数加在一起看作一个数，当前面的数为负数的时候则舍去，将当前这个数作为新的起点。用一个变量Max记录一下最大的值，当最大值被替换的时候记录一下终点和起点。理解了还是很容易的。

#include<bits/stdc++.h>

using namespace std;

const int maxn = 1e5+100;

int num[maxn],dp[maxn];

int Start,End,now_start,Max;

int main()

{

    int t;

    int T;

    scanf("%d",&t);

    T = t;

    while(t--)

    {

        memset(dp,0,sizeof(dp));

        num[0] = -1;

        now_start = 1;

        Max = -1000000;

        int n;

        scanf("%d",&n);

        for(int i=1;i<=n;i++)

            scanf("%d",&num[i]);

        for(int i=1;i<=n;i++)

        {

            if(num[i] + dp[i-1] >= num[i])//前面的数不是负数

                dp[i] = num[i] + dp[i-1];

            else

            {

                dp[i] = num[i];//前面一个数是负数，将现在这个数作为新的起点

                now_start = i;//当前起点，当出现最大值的用得到

            }

            if(dp[i] >= Max)//出现最大值，记录最大值，终点和起点

            {

                Start = now_start;

                End = i;

                Max = dp[i];

            }

        }

        printf("Case %d:\n",T-t);

        if(t != 0)

            printf("%d %d %d\n\n",Max,Start,End);

        else

            printf("%d %d %d\n",Max,Start,End);

    }

    return 0;

}