Max Sum

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 1003
64-bit integer IO format: %I64d Java class name: Main

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

解题：dp入门题！弱菜的成长之路啊！

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <cstdlib>

 #include <vector>

 #include <climits>

 #include <algorithm>

 #include <cmath>

 #define LL long long

 using namespace std;

 int dp[],num[];

 int main(){

     int kase,i,index,ans,n,k = ;

     scanf("%d",&kase);

     while(kase--){

         scanf("%d",&n);

         for(i = ; i <= n; i++)

             scanf("%d",dp+i);

         num[] = ;

         ans = dp[index = ];

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

             if(dp[i] <= dp[i-]+dp[i]){

                 dp[i] = dp[i-]+dp[i];

                 num[i] = num[i-]+;

             }else num[i] = ;

             if(ans < dp[i]) ans = dp[index = i];

         }

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

         printf("%d %d %d\n",ans,index-num[index],index);

         if(kase) putchar('\n');

     }

     return ;

 }

巴特西

BNUOJ 5227 Max Sum

Max Sum

Input

Output

Sample Input

Sample Output

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