uva12298(生成函数)

生成函数的一般应用；

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=;
const long double PI=acos(-1.0);
struct cp{
    long double r,i;
    cp(long double _r=,long double _i=):r(_r),i(_i){}
    cp operator+(const cp&a){
        return cp(r+a.r,i+a.i);
    }
    cp operator-(const cp&a){
        return cp(r-a.r,i-a.i);
    }
    cp operator*(const cp&a){
        return cp(r*a.r-i*a.i,r*a.i+a.r*i);
    }
    void pu(long double _r=,long double _i=){
        r=_r;i=_i;
    }
}wn,w,A[][maxn],tmp[maxn],x,y;
int n,dig[maxn],rev[maxn],L;
void fft(cp a[],int flag){
    for(int i=;i<n;++i)tmp[i]=a[rev[i]];
    for(int i=;i<n;++i)a[i]=tmp[i];
    for(int i=;i<=n;i<<=){
        cp wn(cos(*PI/i),flag*sin(*PI/i));
        for(int k=;k<n;k+=i){
            cp w(,);
            for(int j=k;j<k+i/;++j){
                x=a[j];y=a[j+i/]*w;
                a[j]=x+y;a[j+i/]=x-y;
                w=w*wn;
            }
        }
    }
    if(flag==-)for(int i=;i<n;++i)a[i].r/=n;
}
void solve(cp aa[],cp bb[],int len){
    memset(rev,,sizeof(rev));
    memset(dig,,sizeof(dig));
    for(n=,L=;n<(len<<);n<<=,++L);
    for(int i=;i<n;++i){
        int ll=;
        for(int t=i;t;t>>=)dig[ll++]=t&;
        for(int j=;j<L;++j)rev[i]=(rev[i]<<)|dig[j];
    }
    for(int i=len;i<n;++i)aa[i].pu(),bb[i].pu();
    //for(int i=0;i<n;++i)cout<<aa[i].r<<' ';cout<<endl;
    fft(aa,);fft(bb,);
    for(int i=;i<n;++i)aa[i]=aa[i]*bb[i];
    fft(aa,-);
}
bool is[],ex[][];
char ss[];
int main(){
    for(long long i=;i<;++i){
        if(!is[i]){
            for(long long j=i+i;j<;j+=i)is[j]=;
        }
    }
    int a,b,c,op;
    while(scanf("%d%d%d",&a,&b,&c)==&&(a||b||c)){
        memset(ex,,sizeof(ex));
        for(int i=;i<=c;++i){
            scanf("%d",&op);scanf("%s",ss);
            if(ss[]=='S')ex[][op]=;
            else if(ss[]=='H')ex[][op]=;
            else if(ss[]=='C')ex[][op]=;
            else ex[][op]=;
        }
        for(int i=;i<;++i)
          for(int j=;j<b;++j){
            if(is[j]&&ex[i][j]){
                A[i][j].pu();
            }
            else A[i][j].pu();
        }
        solve(A[],A[],b);
        solve(A[],A[],b);
        solve(A[],A[],b);
        for(int i=a;i<=b;++i){
            printf("%lld\n",(long long)(A[][i].r+0.5));
        }
        cout<<endl;
    }
    //system("pause");
    return ;
}
/*
12 20 2
4S 6H
0 0 0
*/