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hdu 1402 大数A*B模板(FFT)

题目链接

参考博客

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <complex>
#include <iostream>
using namespace std;
typedef long long ll;
template <class T>
void read(T &x){
    char c;
    bool op = 0;
    while(c = getchar(), c < ‘0‘ || c > ‘9‘)
        if(c == ‘-‘) op = 1;
    x = c - ‘0‘;
    while(c = getchar(), c >= ‘0‘ && c <= ‘9‘)
        x = x * 10 + c - ‘0‘;
    if(op) x = -x;
}
template <class T>
void write(T x){
    if(x < 0) putchar(‘-‘), x = -x;
    if(x >= 10) write(x / 10);
    putchar(‘0‘ + x % 10);
}
const int N = 200005;
const double PI = acos(-1);
typedef complex <double> cp;
char sa[N], sb[N];
int n = 1, lena, lenb, res[N];
cp a[N], b[N], omg[N], inv[N];
void init(){
    for(int i = 0; i < n; i++){
        omg[i] = cp(cos(2 * PI * i / n), sin(2 * PI * i / n));
        inv[i] = conj(omg[i]);
    }
}
void fft(cp *a, cp *omg){
    int lim = 0;
    while((1 << lim) < n) lim++;
    for(int i = 0; i < n; i++){
        int t = 0;
        for(int j = 0; j < lim; j++)
            if((i >> j) & 1) t |= (1 << (lim - j - 1));
        if(i < t) swap(a[i], a[t]); // i < t 的限制使得每对点只被交换一次（否则交换两次相当于没交换）
    }
    for(int l = 2; l <= n; l *= 2){
        int m = l / 2;
        for(cp *p = a; p != a + n; p += l)
            for(int i = 0; i < m; i++){
                cp t = omg[n / l * i] * p[i + m];
                p[i + m] = p[i] - t;
                p[i] += t;
            }
    }
}
int main(){
    while(~scanf("%s%s", sa, sb)){
        n = 1;
        lena = strlen(sa), lenb = strlen(sb);
        while(n < lena + lenb) n *= 2;
        for(int i = 0; i < lena; i++)
            a[i].real(sa[lena - 1 - i] - ‘0‘),a[i].imag(0);
        for(int i = 0; i < lenb; i++)
            b[i].real(sb[lenb - 1 - i] - ‘0‘),a[i].imag(0);
        init();
        fft(a, omg);
        fft(b, omg);
        for(int i = 0; i < n; i++)
            a[i] *= b[i];
        fft(a, inv);
        for(int i = 0; i < n; i++){
            res[i] += floor(a[i].real() / n + 0.5);
            res[i + 1] += res[i] / 10;
            res[i] %= 10;
        }
        int m = lena+lenb-1;
        while(res[m]==0 && m > 0){
            m--;
        }
        for(int i = m; i >= 0; i--)
        {
            putchar(‘0‘ + res[i]);
        }
        puts("");
        memset(res,0, sizeof(res));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
    }
    return 0;
}

hdu 1402 大数A*B模板(FFT)

标签：put   ref   cst   mem   限制   algorithm   return   ios   using   

原文地址：https://www.cnblogs.com/hh13579/p/12699862.html