hdoj 1003 Max Sum【搜索】

 

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 138790    Accepted Submission(s): 32220

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

 

#include<stdio.h>int a[100005];int main(){int t, n, i, j = 1, max, pre, st, en;//pre是当前的总和，st是起始位置，en是结束scanf( "%d", &t );while( t -- ){scanf( "%d", &n );pre = 0;int l = 1; //‘l’（小写的L）是当前的pre的起始位置 这个l不能换成st 因为当pre小于0遇到的第一个正数不一定是满足最大的的起始值max = -1001;for( i = 1; i <= n; i ++ ){scanf( "%d", &a[i] );if( pre >= 0 )pre+=a[i];else{pre = a[i];l = i; //这是确定刚开始的地方  （注意这里）}if( pre > max ){ //如果满足条件 改变max en st的值 max = pre;en = i;st = l;}}printf( "Case %d:\n", j++ ); printf( "%d %d %d\n", max, st, en );if( t )printf( "\n" );}return 0;}