Maximum GCD - UVa 11827 最大公约数 读入有坑
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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); }}
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