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# Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 157635    Accepted Submission(s): 36871

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

要注意最大值值可能是负数 ， so 不要用 result = 0 去比较取最大值 ；

看看能不能过 2 -7 7 这组。

（一开始我写的代码只能保证 在首项  非负是成立 ，so WA ）

``` 1 #include<stdio.h>
2 #include<string.h>
3 int N ;
4 int a[100001] ;
5 int dp[100001] ;
6
7 int main()
8 {
9     freopen ( "a.txt" ,"r" , stdin ) ;
10     int T;
11     int i , j ;
12     int cnt = 0 ;
13     int start , end ;
14     int result ;
15     scanf ( "%d" , &T ) ;
16     while ( T-- )
17     {
18         scanf ( "%d" , &N );
19         memset ( dp , 0 , sizeof(dp) );
20         for ( i = 1 ; i <= N ; i++ )
21             scanf ( "%d" , &a[i] ) ;
22
23         for ( i = 1 ; i <= N ; i++ )
24         {
25                 if ( dp[i - 1] + a[i] >= 0 && dp[i - 1] >= 0)
26                     dp[i] = dp[i - 1] + a[i] ;
27                 else
28                 {
29                     dp[i] = a[i] ;
30                 }
31         }
32         result = 1 ;
33         for ( i = 2 ; i <= N ; i++ )
34         {
35             if( dp[result] < dp[i] )
36                 result = i ;
37         }
38         start = dp[result] ;
39         for ( j = result ; j >= 1 ; j-- )
40         {
41             start-= a[j] ;
42             if ( !start )
43                 end = j;
44         }
45         printf ( "Case %d:
%d %d %d
" , cnt + 1 , dp[result] , end , result ) ;
46         cnt++ ;
47         if ( T )
48             printf ( "
" ) ;
49     }
50     return 0;
51 }```
dp
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• 原文地址：https://www.cnblogs.com/get-an-AC-everyday/p/4251144.html