hdu 1003 Max Sum 解题报告

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003

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

简单的动态规划，最大子串和；

 

   /*********   hdu 1003  ************/
  /*********   琴心&剑胆   ************/
  /*********   2011/5/9 ************/

#include<stdio.h>
int main(){
    int T;
    scanf( "%d",&T );
    int M=T;
    while( T-- ){
        int n;
        scanf( "%d",&n );
        int a[n],ts=0,ti=0,s=0,bi=0,bj=n-1,f=0,max=-32767,mi;
        for( int i=0;i<n;++i ){
            scanf( "%d",&a[i] );
            if(max<a[i]){
                max=a[i];
                mi=i;
            }
            if( a[i]<0 )
                f++;
            ts+=a[i];
            if(ts<0){
                ts=0;
                ti=i+1;
            }
            if(s<ts){
                s=ts;
                bj=i;
                bi=ti;
            }
        }
        printf("Case %d:\n",M-T);
        if( f==n )
            printf( "%d %d %d\n",max,mi+1,mi+1 );
        else
            printf( "%d %d %d\n",s,bi+1,bj+1 );
        if( T )
            printf("\n");
    }
}