UVA 12298 (FFT)

题意:给定一个区间[a,b],给定四种花色的纸牌,每张牌上都有一个合数.对于每一个n属于[a,b]四种牌中各取一张和等于n的方案数.然后限制条件是有几种牌遗失了.

裸的fft,找出所有的合数记录下来然后去掉遗失的,对每个花色FFT,然后四个花色的点表达式相乘求出IDFT就是方案数了~

坑点:所有的都要换成long double!!!

#include <cstring>
#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
const long double pi = acos (-1);
#define maxn 1<<19

struct plex {
long double x, y;
plex (long double _x = 0.0, long double _y = 0.0) : x (_x), y (_y) {}
plex operator + (const plex &a) const {
return plex (x+a.x, y+a.y);
}
plex operator - (const plex &a) const {
return plex (x-a.x, y-a.y);
}
plex operator * (const plex &a) const {
return plex (x*a.x-y*a.y, x*a.y+y*a.x);
}
};

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;
}

plex A[1 << 19];
void FFT(plex *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);
plex wm = plex(cos(DFT*2*pi/m), sin(DFT*2*pi/m));
for(int k = 0; k < len; k += m)
{
plex w = plex(1, 0);
for(int j = 0; j < (m >> 1); j++)
{
plex t = w*A[k + j + (m >> 1)];
plex 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].x /= len, A[i].y /= len;
for(int i = 0; i < len; i++) a[i] = A[i];
return;
}

char op[11111][11];
bool is_prime[51111];
int a, b, c, d;
plex S[maxn], H[maxn], D[maxn], C[maxn];

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

int main () {
//freopen ("in.txt", "r", stdin);
init ();
while (scanf("%d %d %d", &a, &b, &c), a || b || c) {
int len = 1;
while(len <= b) len <<= 1;
len <<= 3;
for(int i = 0; i <= b; i++)
if(is_prime[i]) {
S[i] = H[i] = C[i] = D[i] = plex(1, 0);
}
else S[i] = H[i] = C[i] = D[i] = plex(0, 0);
for(int i = b + 1; i < len; i++) S[i] = H[i] = C[i] = D[i] = plex(0, 0);
int num;
char type;
for(int i = 0; i < c; i++)
{
scanf("%d%c", &num, &type);
switch(type)
{
case 'S': S[num] = plex(0, 0); break;
case 'H': H[num] = plex(0, 0); break;
case 'C': C[num] = plex(0, 0); break;
case 'D': D[num] = plex(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", (long long)(S[i].x + 0.5));
putchar('\n');
}
return 0;
}