hdu1003（Max Sum）DP类题目

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

 

题意：这题目是想要说明一串数据中，找出它的最大值，这个最大值得保证是某一个数到另一个数的和。例如：6 -1 5 4 -7，最大值便为14，其计算过程为（6-1+5+4）；0 6 -1 1 -6 7 -5，最大值便为7，其计算过程为（0+6-1+1-6+7）。这是比较简单的DP题，但是这道题最关键的，在于换行的问题：具体的输入输出应该为：

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代码：

#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;
}