杭电ACM1003 Max Sum

最新推荐文章于 2022-04-06 14:55:18 发布

guyeyidian

最新推荐文章于 2022-04-06 14:55:18 发布

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文章标签： input output integer each

本文链接：https://blog.youkuaiyun.com/guyeyidian/article/details/5924188

本文提供了一道杭电ACM 1003题目的详细解答，该题目要求从给定序列中找出具有最大和的子序列，并给出子序列的起始和结束位置。通过分析输入输出格式及示例，文章提供了完整的C语言代码实现。

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

杭电ACM1003

Max Sum

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
  
  
  
  
   
   解法：
  
  
  
  
   
   #include"stdio.h"
int n;
int a[100001];
int left,right;
void input()
{
    int i;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
        scanf("%d",&a[i]);
}
int Max(int a,int b)
{
    return a>b?a:b;
}
int count()
{
    int i,t;
 int max_sum,m;
 max_sum=a[1];
 m=a[1];
 left=1;
    right=1;
 t=1;
    for(i=2;i<=n;i++)
 {
  if(m+a[i]<a[i])
  {
   m=a[i];
   t=i;
  }
  else
   m+=a[i];
  if(max_sum<m)
  {
   max_sum=m;
   left=t;
   right=i;
  }
 }
    return max_sum;
}
int main()
{
    int T,cases=0,s;
    scanf("%d",&T);
    while(T--)
    {
        cases++;
        input();
  s=count();
        printf("Case %d:/n",cases);
  printf("%d %d %d/n",s,left,right);
  if(T>0)printf("/n");
    }
    return 0;
}