HDU1003 DP

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 221949    Accepted Submission(s): 52183

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

题意：

求最大连续子序列和

代码：

//水
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    int t,n;
    scanf("%d",&t);
    for(int k=1;k<=t;k++)
    {
        int dp[100005][3];
        memset(dp,0,sizeof(dp));
        int a[100005];
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
        dp[1][0]=a[1];
        dp[1][1]=1;
        dp[1][2]=1;
        for(int i=2;i<=n;i++)
        {
            if(dp[i-1][0]<0)
            {
                dp[i][0]=a[i];
                dp[i][1]=i;
                dp[i][2]=i;
            }
            else
            {
                dp[i][2]=i;
                dp[i][1]=dp[i-1][1];
                dp[i][0]=a[i]+dp[i-1][0];
            }
        }
        int tem=-200000000,temi;
        for(int i=1;i<=n;i++)
        {
            if(dp[i][0]>tem)
            {tem=dp[i][0];temi=i;}
        }
        printf("Case %d:\n",k);
        printf("%d %d %d\n",dp[temi][0],dp[temi][1],dp[temi][2]);
        if(k!=t) printf("\n");
    }
    return 0;
}