codeforces 145C – Lucky Subsequence mark

145C - Lucky Subsequence

題意：在一串數中任取K個數，問有幾種情況是不含有兩個相同的luky number

組合數計算模版，這個是超大型的組合數啊

#include<cstdio>#include<iostream>#include<map>#define X 100010#define mod 1000000007using namespace std;typedef long long LL;map<int,int> g;map<int,int>::iterator gi;LL f[X]={1},dp[X]={1};LL mi(LL a,LL b){    LL as=1;    while(b){        if(b&1)as=(as*a)%mod;         b>>=1;a =(a *a)%mod;    }    return as;}LL c(LL a,LL b){    if(a<b)return 0;    LL as=f[a];    as=(as*mi(f[  b],mod-2))%mod;    as=(as*mi(f[a-b],mod-2))%mod;    return as;}bool luck(int x){    do    if(x%10!=4&&x%10!=7) return 0;    while(x/=10);    return 1;}int main(void){     for(int i=1;i<X;i++)        f[i]=(f[i-1]*i)%mod;    int n,k,x;    LL cnt=0,ans=0;    scanf("%d%d",&n,&k);    for(int i=1;i<=n;i++)    {        scanf("%d",&x);        if(luck(x))g[x]++;        else cnt++;    }    int pos=1;    for(gi=g.begin();gi!=g.end();gi++)    {        for(int i=min(pos,k);i>0;i--)        {            dp[i]+=gi->second*dp[i-1];            dp[i]%=mod;        }        pos++;    }    pos--;    for(int i=0;i<=pos&&i<=k;i++)    {        ans+=c(cnt,k-i)*dp[i];        ans%=mod;    }    cout<<ans<<endl;    return 0;}