HDU 1003 Max Sum （dp）

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

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main(){
    int t,n,a[100010],cnt=1,dp[100010];
    scanf("%d",&t);
    int q=t;
    while(t--){
        scanf("%d",&n);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            dp[i]=a[i];
        }
        int maxx=dp[1];
        int t=1;
        for(int i=1;i<=n;i++){
            dp[i]=max(dp[i],dp[i-1]+a[i]);
            if(dp[i]>maxx){
                maxx=dp[i];
                t=i; //记录最长子序列的末位置 
            }
        }
        int sum=0,tt;
        for(int i=t;i>=1;i--){
            sum+=a[i];
            if(sum==maxx){ //找最长子序列的初始位置 
                tt=i;
            }
        }
        printf("Case %d:\n",cnt++);
        printf("%d %d %d\n",maxx,tt,t);
        if(q!=(cnt-1))
            printf("\n");
    }
}