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！

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cstdlib>

 5 #include <vector>

 6 #include <climits>

 7 #include <algorithm>

 8 #include <cmath>

 9 #define LL long long

10 using namespace std;

11 int dp[100010],num[100010]; 12 int main(){ 13     int kase,i,index,ans,n,k = 1; 14     scanf("%d",&kase); 15     while(kase--){ 16         scanf("%d",&n); 17         for(i = 1; i <= n; i++) 18             scanf("%d",dp+i); 19         num[1] = 0; 20         ans = dp[index = 1]; 21         for(i = 2; i <= n; i++){ 22             if(dp[i] <= dp[i-1]+dp[i]){ 23                 dp[i] = dp[i-1]+dp[i]; 24                 num[i] = num[i-1]+1; 25             }else num[i] = 0; 26             if(ans < dp[i]) ans = dp[index = i]; 27  } 28         printf("Case %d:\n",k++); 29         printf("%d %d %d\n",ans,index-num[index],index); 30         if(kase) putchar('\n'); 31  } 32     return 0; 33 }

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