BNUOJ 5227 Max Sum

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 ;
 }