HDU 1003 Max Sum【最大连续子序列的和】

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

思路

状态转移方程是：sum[i]=sum[i-1]>0?:sum[i-1]+a[i]:a[i]

源码

#include<stdio.h> #define min -99999999 int main () {     int i, n, m, sum, end, start, j, son, b, maxsum;     scanf("%d", &n) ;     for(i=1; i<=n; i++ )      {         scanf("%d", &m ) ;         start=0;         end=0 ;         maxsum=min ;         sum=0 ;         b=1 ;         for(j=1; j<=m; j++ )          {             scanf("%d", &son) ;             sum+=son ;             if(sum>maxsum)              {                  maxsum=sum ;                  start=b;                  end=j ;             }                      if(sum<0)                       {                             sum = 0;                             b = j+1 ;                      }               }         printf( "Case %d:\n", i) ;         printf( "%d %d %d\n", maxsum, start, end);         if(i!=n)                       printf("\n") ;     } }