hdu_1003_Max Sum_201311271630

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 121261    Accepted Submission(s): 28030

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

 

#include <stdio.h>

int main()
{
    int k,T;
    scanf("%d",&T);
    for(k=1;k<=T;k++)
    {
       int i,j;
       int sta,end,n,m;
       int t=0,max=-1111;
       scanf("%d",&n);
       sta=end=j=1;
       for(i=1;i<=n;i++)
       {
            scanf("%d",&m);
            t += m;
            if(t>max)
            {
                max=t;
                sta=j;
                end=i;
            }
            if(t<0)
            {
                t=0;
                j=i+1;
            }
        }
        if(k-1)
        printf("\n");
        printf("Case %d:\n",k);
        printf("%d %d %d\n",max,sta,end);
    }
    return 0;
}

 

最大连续子序和

不错的资源：

http://blog.csdn.net/luxiaoxun/article/details/7438315

http://blog.csdn.net/code_pang/article/details/7772200

http://blog.csdn.net/shahdza/article/details/6302823

http://www.cnblogs.com/CCBB/archive/2009/04/25/1443455.html

 

参考代码：

#include <stdio.h>
int main()
{int n,T,a,sta,end,max,k,i,p,t;
 
 scanf("%d",&T);
 for(p=1;p<=T;p++) {
  scanf("%d",&n);
  max=-9999;     //因为一个数a 是-1000~1000的，所以这里相当于变成最小值
  t=0;           //表示 某段连续和
  sta=end=k=1;   // sta最大和的开始,end最大和的结束,k记录每次求和的开始
  for(i=1;i<=n;i++) {
   scanf("%d",&a);  
   
   t+=a;        
   if(t>max) {   //记录最大连续和的值
    max=t;
    sta=k;
    end=i;
   }
   if(t<0) {  
    t=0;
    k=i+1;
   }
  }
  
  if(p!=1) printf("/n");
  printf("Case %d:/n",p);
  printf("%d %d %d/n",max,sta,end);
 }
}