Max SumTime 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
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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:\n%d %d %d\n" , cnt + 1 , dp[result] , end , result ) ; 46 cnt++ ; 47 if ( T ) 48 printf ( "\n" ) ; 49 } 50 return 0; 51 }
原文：http://www.cnblogs.com/get-an-AC-everyday/p/4251144.html
