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cf1203C C. Common Divisors

作者:互联网

问题分析

gcd + 优化 (一个数的因数至少有一半小于等于 sqrt(这个因数)

AC代码如下:


#include<iostream>
#include<cstring>
#include<cmath>
#include<set>
using namespace std;
const int maxn=4e5+3;
long long num[maxn] ;
set<long long > number;

long long gcd(long long a,long long b){
    while(b){
    long long t=b;
    b=a%b;
    a=t;
    }
    return a;
}
int main(){
   int n;
  scanf("%lld,%lld",&n,&num[0]);
   long long gc=num[0];
   for(int i=1;i<n;i++){
  scanf("%lld",&num[i]);
     gc=gcd(gc,num[i]);
   }
    long long sum;
        for(int i=1;i*i<=gc;i++){
        if(gc%i==0)
        sum++;
        if(i*i!=gc)
        sum++;
    }
    cout<<sum<<endl;
    return 0;
}

标签:cf1203C,gcd,int,long,num,Common,include,Divisors,lld
来源: https://www.cnblogs.com/maxv/p/11767774.html