Maximum GCD - UVa 11827

Given the N integers, you have to find the maximum GCD (greatest common divisor) of every possiblepair of these integers.InputThe 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) positiveintegers that you have to find the maximum of GCD.OutputFor each test case show the maximum GCD of every possible pair.Sample Input310 20 30 407 5 12125 15 25Sample Output20125

主要是为了练习 set的使用，其实可以暴力两个for求gcd。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<set>
#define ll long long
using namespace std;
const int mx = 1000;
int num[mx];
char str[10000];
set<int>s;
int slove(int tol){
	s.clear();
	int te, qr,ob,ans = 0;
	for(int i = 0; i < tol; i++){
		te = num[i];
		qr = sqrt(te);
		
			
		for(int k = 1; k <= qr; k++)
			if(te%k == 0){
			//	  cout<<"k="<<k<<endl; 
				ob = *s.lower_bound(k);    //必须加星号 
				if(ob == k)
					ans = max(k,ans);
				else
					s.insert(k);
					
				int gg = te/k;
				//	cout<<"gg="<<gg<<endl; 
				ob = *s.lower_bound(gg);    //必须加星号 
				if(ob == gg)
					ans = max(gg,ans);
				else
					s.insert(gg);
			}
			
		
	}
	return ans;
	
}

int main(){
	int n,T,va,tol,len;
	scanf("%d",&T); 
	getchar();
	while(T--){
		gets(str);
		va = 0,tol = 0;
		len = strlen(str);
		for(int i = 0; i < len; i++)
			if(str[i]>='0'&&str[i]<='9')
				va = va*10 + str[i]-'0';
			else{
				num[tol++] = va;
				va = 0;
			}
		if(va != 0)
			num[tol++] = va;
				
		int ans=0;
		ans = slove(tol);
		cout<<ans<<endl;
	}
	
	//cout<<gcd(6,3)<<endl;
	return 0;
}

暴力代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define ll long long
using namespace std;
const int mx = 1000;
int num[mx];
char str[10000];
int gcd(int a, int b){
	int te;
	while(b){
		te = a;
		a = b;
		b = te%b;
	}	
	
	return a;
}

int main(){
	int n,T,va,tol,len;
	scanf("%d",&T); 
	getchar();
	while(T--){
		gets(str);
		va = 0,tol = 0;
		len = strlen(str);
		for(int i = 0; i < len; i++)
			if(str[i]>='0'&&str[i]<='9')
				va = va*10 + str[i]-'0';
			else{
				num[tol++] = va;
				va = 0;
			}
		if(va != 0)
			num[tol++] = va;
				
	/*	for(int i = 0; i < tol; i++){
			cout<<num[i]<<","<<endl;
		}*/	
		int ans=0;
		for(int i = 0; i < tol; i++)
			for(int j = i+1; j < tol; j++)
			 ans = max(ans,gcd(num[i],num[j])); 
		cout<<ans<<endl;
	}
	
	//cout<<gcd(6,3)<<endl;
	return 0;
}