Maximum GCD - UVa 11827 最大公约数 读入有坑

Problem J

Maximum GCD 
Input: Standard Input

Output: Standard Output

                       

Given the N integers, you have to find the maximum GCD(greatest common divisor) of every possible pair of these integers.

 

Input

The first line of input is an integer N(1<N<100) that determines the number of test cases.

The following N lines are the N test cases. Each test case contains M (1<M<100) positive integers that you have to find the maximum of GCD.

 

 

Output

For each test case show the maximum GCD of every possible pair.

 

Sample Input	Output for Sample Input
3 10 20 30 40 7  5 12 125 15 25	20 1 25

题意：求出每组数中每两个数最大的最大公约数。

思路：读入可能有多余空格……被坑了几次……

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int mi(int a,int b)
{ if(b==0)
   return a;
  else
   return mi(b,a%b);
}
bool cmp(int a,int b)
{ return a>b;
}
int num[101000];
char c,d;
int main()
{ int t,i,j,k,pos,ans,ret;
  scanf("%d%c",&t,&d);
  c=' ';
  while(t--)
  { c=' ';
    pos=1;
    ans=0;
    while(c!='\n')
    { scanf("%c",&c);
      k=c-'0';
      if(k>=0 && k<=9)
       num[pos]=num[pos]*10+k;
      else if(c==' ' && num[pos]>0)
       pos++;
    }
    if(num[pos]==0)
     pos--;
    sort(num+1,num+1+pos,cmp);
    for(i=1;i<pos;i++)
    { if(ans>=num[i])
       break;
      for(j=i+1;j<=pos;j++)
      { if(ans>=num[j])
         break;
        else
        ret=mi(num[i],num[j]);
        if(ret>ans)
         ans=ret;
      }
    }
    printf("%d\n",ans);
  }
}