HDU1003 Max Sum

                                                                 Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 112106    Accepted Submission(s): 25891

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


解题思路：求和最大连续子串。。。。。。。。。用max记录当前最大和，一旦sum小于0，则置0从新累加。

#include<stdio.h> int main() {     int a[100002];     int t,n;     int sum;     int max1,maxx,maxy;     int i,j,k;     scanf("%d",&t);     for(k=1;k<=t;k++)     {         max1=-999999999;         sum=0;         scanf("%d",&n);         for(i=0;i<n;i++)             scanf("%d",&a[i]);         j=0;         for(i=0;i<n;i++)         {             //printf("\ni=%d",i);             sum+=a[i];             // printf("%d ",sum);             if(sum>max1)             {                 max1=sum;                 maxx=j+1;                 maxy=i+1;                 //printf("-------%d %d %d\n",max1,maxx,maxy);             }             if(sum<0)             {                 sum=0;                 j=i+1;             }         }         if(k!=1)             printf("\n");         printf("Case %d:\n",k);         printf("%d %d %d\n",max1,maxx,maxy);     }     return 0; }