【洛谷 P2553】 [AHOI2001]多项式乘法（FFT）

题目链接
简单处理一下输入，\(fft\)模板题。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#define re register
using namespace std;
const int MAXN = 1000010;
const double PI = M_PI;
struct complex{
    double x, y;
    complex(double xx = 0, double yy = 0){ x = xx; y = yy; }
}a[MAXN], b[MAXN], *f;
inline complex operator + (complex a, complex b){
    return complex(a.x + b.x, a.y + b.y);
}
inline complex operator - (complex a, complex b){
    return complex(a.x - b.x, a.y - b.y);
}
inline complex operator * (complex a, complex b){
    return complex(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
}
inline int read(){
    re int s = 0, w = 1;
    re char ch = getchar();
    while(ch < '0' || ch > '9'){ ch = getchar(); if(ch == '-') w = -1; }
    while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }
    return s * w;
}
int r[MAXN], n, m;
void FFT(complex *f, int mode){
    for(re int i = 0; i < n; ++i) if(i < r[i]) swap(f[i], f[r[i]]);
    for(re int p = 2; p <= n; p <<= 1){
       re int len = p >> 1;
       re complex tmp(cos(PI / len), mode * sin(PI / len));
       for(re int l = 0; l < n; l += p){
          re complex w(1, 0);
          for(re int k = l; k < l + len; ++k){
             re complex t = w * f[len + k];
             f[len + k] = f[k] - t;
             f[k] = f[k] + t;
             w = w * tmp;
          }
       }
    }
}
inline double d(double x){
    if(fabs(x) < 1e-9) return 0;
    return x;
}
char s[MAXN];
int *len = &n, now, flag, tmp;
int main(){
    while(scanf("%s", s + 1) != EOF){
    int leng = strlen(s + 1); n = m = flag = 0; f = a; len = &n;
    for(int i = 1; i <= leng; ++i){
        if(s[i] >= '0' && s[i] <= '9') now = now * 10 + s[i] - '0';
        if(s[i] == '^'){ flag = 1; tmp = now; now = 0; }
        if(s[i] == '+' || s[i] == ')'){
          if(flag) f[now].x = tmp, *len = max(*len, now); 
          else f[0].x = now;
          flag = now = tmp = 0;
        }
        if(s[i] == '*') f = b, len = &m;
    }
    for(m += n, n = 1; n <= m; n <<= 1);
    for(re int i = 1; i < n; ++i) r[i] = r[i >> 1] >> 1 | ((i & 1) * (n >> 1));
    FFT(a, 1); FFT(b, 1);
    for(re int i = 0; i < n; ++i) a[i] = a[i] * b[i];
    FFT(a, -1);
    for(int i = m; ~i; --i) 
       if(a[i].x > 1e-9){
          if(flag) printf("+");
          printf("%.0f", a[i].x / n, i);
          if(i) printf("a^%d", i);
          flag = 1;
       }
    printf("\n");
    for(int i = 0; i < n; ++i) a[i].x = a[i].y = b[i].x = b[i].y = 0;
    }
    return 0;
}