bzoj4807 -- 组合数

容易看出答案就是C(n,m)。。。

然后高精乘一下就可以了。

对n!分解质因数时，质数p的出现次数是n/p+n/p^2+n/p^3+...

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 101
inline int Min(int x,int y){return x<y?x:y;}
inline int Max(int x,int y){return x<y?y:x;} 
struct bign  
{  
    int len, s[MAXN];  
    bign ()  
    {  
        memset(s, 0, sizeof(s));  
        len = 1;  
    }  
    bign (int num) { *this = num; }  
    bign (const char *num) { *this = num; }  
    bign operator = (const int num)  
    {  
        char s[MAXN];  
        sprintf(s, "%d", num);  
        *this = s;  
        return *this;  
    }  
    bign operator = (const char *num)  
    {  
        for(int i = 0; num[i] == '0'; num++) ;
        len = strlen(num);  
        for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';  
        return *this;  
    }
    void clean()  
    {
        if(len>50)len=50;
        while(len > 1 && !s[len-1]) len--;  
    }
    bign operator * (const bign &b)  
    {  
        bign c;  
        c.len = len + b.len;  
        for(int i = 0; i < len; i++)  
        {  
            for(int j = 0; j < b.len; j++)  
            {  
                c.s[i+j] += s[i] * b.s[j];  
            }  
        }  
        for(int i = 0; i < c.len; i++)  
        {  
            c.s[i+1] += c.s[i]/10;  
            c.s[i] %= 10;  
        }  
        c.clean();  
        return c;  
    }  
    bign operator *= (const bign &b)  
    {  
        *this = *this * b;  
        return *this;  
    }  
}n;  
int i,k,N,M,P[1000010],tot,p[1000010];
bool b[1000010];
long long j;
int main()
{
    scanf("%d%d",&N,&M);
    if(N<M)swap(N,M);
    for(i=2;i<=N;i++){
        if(!b[i])P[++tot]=i;
        for(j=1;P[j]*i<=N&&j<=tot;j++){
            b[P[j]*i]=1;
            if(!(i%P[j]))break;
        }
    }
    for(i=1;i<=tot;i++)
    for(j=P[i];j<=N;j=j*P[i])p[i]+=N/j;
    for(i=1;i<=tot;i++)
    for(j=P[i];j<=M;j=j*P[i])p[i]-=M/j;
    for(i=1;i<=tot;i++)
    for(j=P[i];j<=N-M;j=j*P[i])p[i]-=(N-M)/j;
    n=1;
    for(i=1;i<=tot;i++)
    for(j=0;j<p[i];j++)n*=P[i];
    for(i=n.len-1;i>=0;i--)putchar(n.s[i]+48);
    return 0;
}