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=;k<=T;k++)

     {

        int i,j;

        int sta,end,n,m;

        int t=,max=-;

        scanf("%d",&n);

        sta=end=j=;

        for(i=;i<=n;i++)

        {

             scanf("%d",&m);

             t += m;

             if(t>max)

             {

                 max=t;

                 sta=j;

                 end=i;

             }

             if(t<)

             {

                 t=;

                 j=i+;

             }

         }

         if(k-)

         printf("\n");

         printf("Case %d:\n",k);

         printf("%d %d %d\n",max,sta,end);

     }

     return ;

 }

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=;p<=T;p++) {

   scanf("%d",&n);

   max=-;     //因为一个数a 是-1000~1000的，所以这里相当于变成最小值

   t=;           //表示 某段连续和

   sta=end=k=;   // 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=;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<) {

     t=;

     k=i+;

    }

   }

   if(p!=) printf("/n");

   printf("Case %d:/n",p);

   printf("%d %d %d/n",max,sta,end);

  }

 }

巴特西

hdu_1003_Max Sum_201311271630

Max Sum

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