UVA 11827Maximum GCD

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F - Maximum GCD

Time Limit:1000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Given the N integers, you have to nd the maximum GCD (greatest common divisor) of every possible
pair of these integers.
Input
The rst 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 nd the maximum of GCD.
Output
For each test case show the maximum GCD of every possible pair.
Sample Input
3
10 20 30 40
7 5 12
125 15 25
Sample Output
20
1
25

题目很水暴力就能过难点在如何输入没有停止的数

#include <iostream>#include <string.h>#include <stdio.h>#include <math.h>#include <stdlib.h>#include <sstream>using namespace std;int gcd (int a,int b){    if(b==0)    return a;    else    return gcd(b,a%b);}int main(){    int i,j,n,t,a[105];    cin>>t;    getchar();    while(t--)    {        //getchar();        memset(a,0,sizeof(a));        string str;        getline(cin,str);        stringstream stream(str);        int n=0;        while(stream>>a[n])        {            n++;        }        int ans=1;        for(i=0;i<n;i++)        for(j=i+1;j<n;j++)        ans=max(gcd(a[i],a[j]),ans);        cout<<ans<<endl;    }    return 0;}

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