Light OJ 1067 Combinations(逆元）

来源：互联网发布：淘宝小号怎么养编辑：程序博客网时间：2024/05/18 21:50

Description

Given n different objects, you want to take k of them. How many ways to can do it?

For example, say there are 4 items; you want to take 2 of them. So, you can do it 6 ways.

Take 1, 2

Take 1, 3

Take 1, 4

Take 2, 3

Take 2, 4

Take 3, 4

Input

Input starts with an integer T (≤ 2000), denoting the number of test cases.

Each test case contains two integers n (1 ≤ n ≤ 10⁶), k (0 ≤ k ≤ n).

Output

For each case, output the case number and the desired value. Since the result can be very large, you have to print the result modulo 1000003.

Sample Input

4 2

5 0

6 4

Sample Output

Case 1: 6

Case 2: 1

Case 3: 15

题意：就是求C(n,m)%mod;

分析：这道题得逆元来求，由于mod是质数，我们转化成扩展欧几里德和费马小定理求

1：要求b%mod的逆元，只需exgcd(b,mod,x,y)带入，因为同时在两边模mod得bx=1(%mod)所以x就是

b%mod的逆元。

2：要求b%mod的逆元，直接费马小定理，b'=b^(mod-2)%mod,所以可以直接快速幂来求一下就好了。

代码一：

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<set>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int INF=0x3f3f3f3f;typedef long long LL;const int mod=1e6+3;const int maxn=1e6+10;LL fac[maxn],inv[maxn];int n;void exgcd(LL a,LL b,LL &xx,LL &yy){    if(b==0)    {        xx=1;yy=0;        return;    }    exgcd(b,a%b,xx,yy);    LL t=xx;xx=yy;yy=t-a/b*yy;    return;}int main(){    LL x,y;    fac[0]=1;inv[0]=1;    for(LL i=1;i<=1000000;i++)    {        fac[i]=fac[i-1]*i%mod;        exgcd(fac[i],mod,x,y);        inv[i]=(x%mod+mod)%mod;    }    int cas=1;int a,b;    scanf("%d",&n);    while(n--)    {        scanf("%d%d",&a,&b);        LL ans=fac[a]*inv[b]%mod*inv[a-b]%mod;        printf("Case %d: %lld\n",cas++,ans);    }    return 0;}

代码二：

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<set>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int INF=0x3f3f3f3f;typedef long long LL;const int mod=1e6+3;const int maxn=1e6+100;LL fac[maxn],inv[maxn];LL Pow(LL a,LL b){    LL ans=1;    while(b)    {        if(b&1)            ans=(ans*a)%mod;        a=(a*a)%mod;        b>>=1;    }    return ans;}int main(){    int cas=1;    int n,a,b;    fac[0]=inv[0]=1;    for(int i=1;i<=1000000;i++)    {        fac[i]=(fac[i-1]*i)%mod;        inv[i]=Pow(fac[i],mod-2);    }    scanf("%d",&n);    while(n--)    {        scanf("%d%d",&a,&b);        LL ans=fac[a]*inv[b]%mod*inv[a-b]%mod;        printf("Case %d: %lld\n",cas++,ans);    }    return 0;}

0 0