【FFT】 UVA 12298 Super Poker II

通道：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=84075#problem/C

题意：

就是现在有一堆扑克里面的牌有无数张, 每种合数的牌有4中不同花色各一张(0, 1都不是合数), 没有质数或者大小是0或者1的牌

现在这堆牌中缺失了其中的 c 张牌, 告诉你a, b, c接下来c张不同的丢失的牌, 然后求从这堆牌中拿出各种花色的牌各一张, 得到的点数和是k的种数有多少种(一种组合算作一种), 需要全部所有的a <= k <= b的k对应的结果

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

const int MAX_N = 100007; 
const int MAX_M = 100007;
const long double PI = acos(-1.0);

struct Complex {
    long double r, i;
    Complex(long double _r, long double _i) {
        r = _r;
        i = _i;
    }
    Complex operator + (const Complex &c) {
        return Complex(c.r + r, c.i + i);
    }
    Complex operator - (const Complex &c) {
        return Complex(r - c.r, i - c.i);
    }
    Complex operator * (const Complex &c) {
        return Complex(c.r * r - c.i * i, c.r * i + c.i * r);
    }
    Complex operator / (const int &c) {
        return Complex(r / c, i / c);
    }
    Complex(){}
};
namespace FFT {
    int rev(int id, int len) {
        int ret = 0;
        for(int i = 0; (1 << i) < len; ++i) {
            ret <<= 1;
            if(id & (1 << i)) ret |= 1;
        }
        return ret;
    }
    Complex A[MAX_M << 3];
    void FFT(Complex *a, int len, int DFT) {
        for(int i = 0; i < len; ++i) A[rev(i, len)] = a[i];
        for(int s = 1; (1 << s) <= len; ++s) {
            int m = (1 << s);
            Complex wm = Complex(cos(PI * DFT * 2 / m), sin(PI * DFT * 2 / m));
            for(int k = 0; k < len; k += m) {
                Complex w = Complex(1, 0);
                for(int j = 0; j < (m >> 1); j++) {
                    Complex t = w * A[k + j + (m >> 1)];
                    Complex u = A[k + j];
                    A[k + j] = u + t;
                    A[k + j + (m >> 1)] = u - t;
                    w = w * wm;
                }
            }
        }
        if(DFT == -1) for(int i = 0; i < len; ++i) A[i] = A[i] / len;
        for(int i = 0; i < len; i++) a[i] = A[i];
    }
};

Complex S[MAX_M << 3], H[MAX_M << 3], C[MAX_M << 3], D[MAX_M << 3];
bool prime[MAX_N >> 1];

void init() {
    memset(prime, 1, sizeof prime);
    int limit = MAX_N >> 1;
    for (int i = 2; i < limit; ++i) {
        if (prime[i]) {
            for (int j = 2 * i; j < limit; j += i)
                prime[j] = false;
        }
    }
}

int main() {
    init();
    int a, b, c;
    while (3 == scanf("%d%d%d", &a, &b, &c)) {
        if (0 == a && 0 == b && 0 == c) break;
        for (int i = 0; i <= b; ++i) if (!prime[i]) 
            S[i] = H[i] = C[i] = D[i] = Complex(1, 0);
        else S[i] = H[i] = C[i] = D[i] = Complex(0, 0);
        ++b;
        int len = 1;
        while (len < b) len <<= 1; len <<= 3;
        for (int i = b; i < len; ++i) S[i] = H[i] = C[i] = D[i] = Complex(0, 0); 
        for(int i = 0; i < c; i++) {  
            int v;
            char type;
            scanf("%d%c", &v, &type);  
            switch(type) {  
                case 'S': S[v] = Complex(0, 0); break;  
                case 'H': H[v] = Complex(0, 0); break;  
                case 'C': C[v] = Complex(0, 0); break;  
                case 'D': D[v] = Complex(0, 0); break;  
            }  
        }  
        FFT::FFT(S, len, 1); FFT::FFT(H, len, 1); FFT::FFT(C, len, 1); FFT::FFT(D, len, 1);
        for (int i = 0; i < len; ++i) S[i] = S[i] * H[i] * C[i] * D[i];
        FFT::FFT(S, len, -1);
        for (int i = a; i < b; ++i) printf("%lld\n", (ll)(S[i].r + 0.5));  puts("");
    }
    return 0;
}