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Max Sum
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 |
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 else28 {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 }
Max Sum
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