阶乘的整数分解 fzu1753

题意：给T个组合数的最大公约数。

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 <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <fstream>#include <numeric>#include <iomanip>#include <bitset>#include <list>#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/i;j++)        {            vis[prime[j]*i]=1;            if(i%prime[j]==0)                break;        }    }}void Getprime(){   for(int i=2;i<1e5+5;i++)    vis[i]=true;   vis[0]=0;   vis[1]=0;    for(int i = 2 ; i < 1e5+5 ; i++){        if(vis[i])            for(int j = i + i ; j < 1e5+5 ; j += i){                if(vis[i])                    vis[j] = false;            }    } cnt=0;    for(int i = 0 ; i < 1e5+5 ; i++)        if(vis[i])            prime[cnt++] = i;}int cal(int n,int p){    int sum=0;    while(n)    {        sum+=n/p;        n=n/p;    }    return sum;}long long int mymul(long long int a,long long int b){    long long int res=0;    while(b)    {        if(b&1)            res+=a;        a<<=1;        b>>=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 tmp=cal(n[i],prime[j]);        tmp-=cal(m[i],prime[j]);        tmp-=cal(n[i]-m[i],prime[j]);        res[j]=min(tmp,res[j]);        }        REP(k,0,res[j])        ans*=(LL)prime[j];        }cout<<ans<<endl;}return 0;}