HDU1003——DP——Max Sum

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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很简单的dp，dp[i]表示以i为开头的值

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

const int maxn = 100010;

const int inf = 0x3f3f3f3f;

int a[maxn],dp[maxn];

int T,n;



int main()

{

    scanf("%d",&T);

    for(int cas = 1; cas <= T; cas++){

        scanf("%d",&n);

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

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

            dp[i] = a[i];

        }

        for(int i = n-1; i >= 1; i--){

            dp[i] = max(dp[i], dp[i+1] + a[i]);

        }

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

       //     printf("%d ",dp[i]);

        int index;

        int max1 = -inf;

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

            if(max1 < dp[i]){

                 max1 = dp[i];

                index = i;

            }

        }

        int res = 0;

        int index1;

        for(int i = index; i <= n; i++){

            res += a[i];

            if(res == max1){

                index1 = i;

                break;

            }

        }

    printf("Case %d:\n",cas);

    printf("%d %d %d\n",max1,index,index1);

   if(cas < T) printf("\n");

    }

    return 0;

}