CSU1383-A Easy Problem-模拟

U: A Easy Problem

Description

This is a classical problem. Your job is to calculate the combination formula C(n,m) , and C(n,m) fit a 64-bit unsigned integer.（注意答案的范围）

Input

The first line of input there is one integer T(T≤100), giving the number of test cases in the input. Each test case contains a line with two integer n and m(0≤m≤n<264).

Output

For each case, output the answer C(n,m) corresponding to the input.

Sample Input

14 2

Sample Output

6

这个题目数据是蛮极限的，有点恐怖。

首先想到的是递推，用公式C(n,m)=C(n-1,m)+C(n-1,m-1).但是输入可能有2^63，那肯定就爆了啊

其次就直接先算分子，然后算分母，再除掉，但是想想就知道这样很容易爆掉

所以就先和我们实际算组合数一样写出分子分母，然后约分到分母全被约去，然后将分子乘起来就行

#include <cstdio>#include <iostream>#include <algorithm>#define LL long long#define ULL unsigned long long#define mem(a,n) memset(a,n,sizeof(a))#define fread freopen("in.txt","r",stdin)#define fwrite freopen("out.txt","w",stdout)#define N 10100#define INF 0x3f3f3f3f#define eps 1e-9 using namespace std;ULL gcd(ULL a,ULL b){    return (b==0?a:gcd(b,a%b));}ULL nume[20000],demo[10000];int main(){    ULL a,b,c,t,cnt1,cnt2,i,j;    cin>>t;    while(t--){        cin>>a>>b;        c=1;        cnt1=cnt2=0;        if(a==b||b==0){            cout<<1<<endl;            continue;        }        if(b==a-1||b==1){            cout<<a<<endl;            continue;         }         b=min(b,a-b);         for(i=a-b+1;i<=a;++i){            nume[cnt1++]=i;        }        for(i=2;i<=b;++i){            demo[cnt2++]=i;        }        for(i=0;i<cnt2;++i){            for(j=0;demo[i]!=1&&j<cnt1;++j){                ULL n=gcd(demo[i],nume[j]);                demo[i]/=n;                nume[j]/=n;            }        }        for(int i=0;i<cnt1;++i){            c*=nume[i];        }        cout<<c<<endl;    }    return 0;}