hdu 1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 180180    Accepted Submission(s): 42082

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

 

Author

Ignatius.L

 

 

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一道很简单的dp，以前也写过很多次，一直是看着别人的代码写的，可是一直觉得自己对dp的理解不够深刻，每次不能独立的想出代码，于是决定把dp 的题再重新写一遍，加深自己的理解。。。。

 

题意：求一串数列中最大序列之和，并输出起始位置和结束位置，尽量保证序列最长（即如第二个例子：第一个数字为0，最大值可加可不加，但是就要从0开始计算起始位置）。

 

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int dp[100001],a[100001];  //a[]数组记录每个数字，dp[]数组记录到当前序列的最大序列和
int main()
{
    int n,m,i,j,t=1;
    scanf("%d",&n);
    while(n--)
    {
        memset(dp,0,sizeof(dp)); //dp 全部初始化为0
        scanf("%d",&m);
        for(i=1; i<=m; i++)
            scanf("%d",&a[i]);
        dp[1]=a[1];
        for(i=2; i<=m; i++)
        {
            if(dp[i-1]<0) dp[i]=a[i];   //  若前列的数加起来之和小于0，则忽略前面的值
            else dp[i]=dp[i-1]+a[i];
        }
        int max=dp[1],e=1,f;
        for(i=2; i<=m; i++)
        {
            if(max<dp[i])
            {
                max=dp[i];   //找到最大的序列之和
                e=i;          // 记录末尾加到的数字
            }
        }
        int s=0;
        for(i=e; i>=1; i--)
        {
            s+=a[i];
            if(s==max)
                f=i;     //找到从哪个数开始的位置
        }
        printf("Case %d:\n",t++);
        printf("%d %d %d\n",max,f,e);
        if(n)
            printf("\n");
    }
    return 0;
}