UVA 12298 Super Poker II(FFT+母函数)

Posted 2020-11-06 Heey

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题目大意：

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

 

 

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做法：FFT

技巧：
我们可以把它变成多项式，根据卷积性质，求得计数。
用x^k前的系数表示大小为k的牌有还是没有(1还是0), 一共4个多项式相乘即可, 得到结果的多项式中x^k前的系数代表的就是可以的方案数
因为我们用卷积（0,1,2,0多项式0是第一项），4个多项式相乘后，第len个位置的值是从4个和为len的地方来

所以可以完成统计a<->b的和的数量

 注意精度问题

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 附上代码：

 

#include<bits/stdc++.h> 
using namespace std;
const double eps(1e-8);
typedef long long lint;

const long double PI = acos(-1.0);

bool isPrime[50010];
bool check[50010];

void init()
{
    memset(isPrime, 0, sizeof(isPrime));
    memset(check, 0, sizeof(check));
    isPrime[0] = isPrime[1] = 1;
    check[0] = check[1] = 1;
    for(int i = 2; i <= 50000; i++)
    {
        if(check[i]) continue;
        isPrime[i] = 1;
        for(int j = i; j <= 50000; j += i)
            check[j] = 1;
    }
    return;
}

struct Complex
{
    long double real, image;
    Complex(long double _real, long double _image)
    {
        real = _real;
        image = _image;
    }
    Complex(){}
};

Complex operator + (const Complex &c1, const Complex &c2)
{
    return Complex(c1.real + c2.real, c1.image + c2.image);
}

Complex operator - (const Complex &c1, const Complex &c2)
{
    return Complex(c1.real - c2.real, c1.image - c2.image);
}

Complex operator * (const Complex &c1, const Complex &c2)
{
    return Complex(c1.real*c2.real - c1.image*c2.image, c1.real*c2.image + c1.image*c2.real);
}

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[1 << 19];
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(DFT*2*PI/m), sin(DFT*2*PI/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].real /= len, A[i].image /= len;
    for(int i = 0; i < len; i++) a[i] = A[i];
    return;
}

Complex S[1 << 19], H[1 << 19], C[1 << 19], D[1 << 19];

int main()
{
    int a, b, c;
    init();
    while(scanf("%d %d %d", &a, &b, &c), a || b || c)
    {
        int len = 1;
        while(len <= b) len <<= 1;
        len <<= 3;//好像因为有4个多项式 就要多开一点 
        for(int i = 0; i <= b; i++)
            if(!isPrime[i]) S[i] = H[i] = C[i] = D[i] = Complex(1, 0);
            else S[i] = H[i] = C[i] = D[i] = Complex(0, 0);
        for(int i = b + 1; i < len; i++) S[i] = H[i] = C[i] = D[i] = Complex(0, 0);
        int num;
        char type;
        for(int i = 0; i < c; i++)
        {
            scanf("%d%c", &num, &type);
            switch(type)
            {
                case ‘S‘: S[num] = Complex(0, 0); break;
                case ‘H‘: H[num] = Complex(0, 0); break;
                case ‘C‘: C[num] = Complex(0, 0); break;
                case ‘D‘: D[num] = Complex(0, 0); break;
            }
        }
        FFT(S, len, 1); FFT(H, len, 1); FFT(C, len, 1); FFT(D, len, 1);
        for(int i = 0; i < len; i++)
            S[i] = S[i]*H[i]*C[i]*D[i];
        FFT(S, len, -1);
        for(int i = a; i <= b; i++)
            printf("%lld\n", (lint)(S[i].real + 0.5));
        putchar(‘\n‘);
    }
    return 0;
}

 

 

 

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