Codeforces Round #422 (Div. 2) A. I'm bored with life

题目大意

给出两个数，计算阶乘的gcd。

题解

阶乘gcd其实就是较小数的阶乘。
复杂度o(min(a,b)).

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int read(){    char ch=getchar();int f=0;    while(ch<'0'||ch>'9') ch=getchar();    while(ch>='0'&&ch<='9')    {f=(f<<1)+(f<<3)+ch-'0';    ch=getchar();}    return f;}int gcd(int a,int b){    return !b?a:gcd(b,a%b);}int main(){    int a=read(),b=read(),x,s=1;    x=min(a,b);    int ans=1;    for(int i=1;i<=x;i++)    s*=i;    cout<<s;}