BestCoder Round #60 HDU5504 GT and sequence-优快云博客

本文链接：https://blog.youkuaiyun.com/qq1319479809/article/details/49228493

本文介绍了一道编程竞赛题目，目标是在给定整数序列中选择若干数（至少一个），使得这些数的乘积尽可能大。文章提供了问题描述、输入输出样例及解析，并附带源代码实现。

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GT and sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 571 Accepted Submission(s): 129

Problem Description

You are given a sequence of

N integers.

You should choose some numbers(at least one),and make the product of them as big as possible.

It guaranteed that **the absolute value of** any product of the numbers you choose in the initial sequence will not bigger than

263−1 .

Input

In the first line there is a number

T (test numbers).

For each test,in the first line there is a number

N ,and in the next line there are

N numbers.

1≤T≤1000

1≤N≤62

You'd better print the enter in the last line when you hack others.

You'd better not print space in the last of each line when you hack others.

Output

For each test case,output the answer.

Sample Input

Sample Output

Source

BestCoder Round #60

出题人：

注意先特判 $0$ 的情况：如果读入的数据有 $0$ ，

那么去掉所有的 $0$ 且最后答案和 $0$ 取一个max。

剩下的正数显然全部乘起来比较优。对于负数的话，

如果个数是奇数个我们就去掉绝对值最小的那一个，然后全部乘起来即可。

当时SB了，没考遇到0的情况。以后要仔细读题，弄懂题意再敲代码！！！！！

#include
  
   
#include
   
    
#include
    
     
#include 
     
      
#include
      
       
#include
       
         using namespace std; long long a[120]; int main() { long long sumd,x,zero,sumx,maxn; int i,t,n,flag; scanf("%d",&t); while(t--) { sumx=sumd=1; maxn=0; zero=flag=0; scanf("%d",&n); for(i=0; i
        
         0)sumd*=x; if(x<0) { flag++; if(flag==1) maxn=x; maxn=max(x,maxn); sumx*=x; } if(x==0) zero++; } if(n==1) { printf("%lld\n",x); continue; } if(flag%2) sumx/=maxn; if(zero==n-1&&maxn<0||zero==n) sumx=0; printf("%lld\n",sumx*sumd); } return 0; }