题意：给T个组合数的最大公约数，
阶乘的整数分解 fzu1753
。将每个组合数的素数分解式求出来，把每个素数的最小次数乘起来就是最大公约数。组合数可以写成阶乘的形式，然后利用阶乘的整数分解就可以得到组合数的整数分解。
代码：
#include<cstdlib>#include<cctype>#include<cstring>#include<cstdio>#include<cmath>#include<climits>#include #include<vector>#include<string>#include<iostream>#include<sstream>#include<map>#include<set>#include<queue>#include<stack>#include<fstream>#include<numeric>#include<iomanip>#include<b>#include<li>#include<stdexcept>#include<functional>#include<utility>#include<ctime>using namespace std;#define PB push_back#define MP make_pair#define REP(i,x,n) for(int i=x;i<(n);++i)#define FOR(i,l,h) for(int i=(l);i<=(h);++i)#define FORD(i,h,l) for(int i=(h);i>=(l);--i)#define SZ(X) ((int)(X).size())#define ALL(X) (X).begin(), (X).end()#define RI(X) scanf("%d", &(X))#define RII(X, Y) scanf("%d%d", &(X), &(Y))#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))#define DRI(X) int (X); scanf("%d", &X)#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)#define OI(X) printf("%d",X);#define RS(X) scanf("%s", (X))#define MS0(X) memset((X), 0, sizeof((X)))#define MS1(X) memset((X), -1, sizeof((X)))#define LEN(X) strlen(X)#define F first#define S second#define Swap(a, b) (a ^= b, b ^= a, a ^= b)#define Dpoint strcut node{int x,y}#define cmpd int cmp(const int &a,const int &b){return a>b;} /*#ifdef HOME freopen("in.txt","r",stdin); #endif*/const int MOD = 1e9+7;typedef vector<int>VI;typedef vector<string>VS;typedef vector<double>VD;typedef long long LL;typedef pair<int,int>PII;//#define HOMEint Scan(){ int res = 0, ch, flag = 0; if((ch = getchar()) == '-') //判断正负 flag = 1; else if(ch >= '0' && ch <= '9') //得到完整的数 res = ch - '0'; while((ch = getchar()) >= '0' && ch <= '9' ) res = res * 10 + ch - '0'; return flag ? -res : res;}/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/long long int prime[100000+5];int vis[100000+5];int cnt;void getprime(){ cnt=0; for(int i=2;i<=100000;i++) if(!vis[i]) { prime[cnt++]=i; for(int j=0;j<cnt&&prime[j]<=100000 5="" b="" cnt="0;" i="2;i<1e5+5;i++)" int="" j="i" long="" n="n/p;" res="0;" return="" sum="0;" void="">>=1; } return res;}long long int mypow(long long int a,long long int b){ long long int res=1; while(b) { if(b&1) res=mymul(res,a); a=mymul(a,a); b>>=1; } return res;}int res[101000];int main(){int t;Getprime();while(RI(t)!=EOF){ int n[160],m[160]; REP(i,0,cnt) res[i]=INT_MAX; int bound=INT_MAX; REP(i,0,t) { RII(n[i],m[i]); bound=min(bound,n[i]);} long long int ans=1; for(int j=0;prime[j]<=bound;j++) { for(int i=0;i<t;i++) int="" pre="" return="" tmp="cal(n[i],prime[j]);" tmp-="cal(m[i],prime[j]);"></p><p> 版权声明：本文为博主原创文章，未经博主允许不得转载，电脑资料
《阶乘的整数分解 fzu1753》(https://www.unjs.com)。</p></t;i++)></cnt&&prime[j]<=100000></int,int></double></string></int></ctime></utility></functional></stdexcept></list></bitset></iomanip></numeric></fstream></stack></queue></set></map></sstream></iostream></string></vector></climits></cmath></cstdio></cstring></cctype></cstdlib>