hdu1402(大数a*b&fft模板)

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1402

题意: 给出两个长度1e5以内的大数a, b, 输出 a * b.

思路: fft模板

详情参见:  m.blog.csdn.net/f_zyj/article/details/76037583

　　　　   http://blog.csdn.net/sdj222555/article/details/9786527

　　　　  https://wenku.baidu.com/view/8bfb0bd476a20029bd642d85.html

可以将 a, b 看成两个多项式, 每个数位为一项, 每一位上的数字即为所在项的系数.所以直接fft即可.

代码:

#include <iostream>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <stdio.h>
using namespace std;

const double PI = acos(-1.0);

struct Complex{//复数结构体
    double x, y;//实部,虚部
    Complex(double _x = 0.0, double _y = 0.0){
        x = _x;
        y = _y;
    }
    Complex operator -(const Complex &b) const{
        return Complex(x - b.x, y - b.y);
    }
    Complex operator +(const Complex &b) const{
        return Complex(x + b.x, y + b.y);
    }
    Complex operator *(const Complex &b) const{
        return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
    }
};

//进行FFT和IFFT反转变化
//位置i和(i二进制反转后位置)互换
void change(Complex y[], int len){//len必须为2的幂
    for(int i = 1, j = len / 2; i < len - 1; i++){
        if(i < j) swap(y[i], y[j]); //交换互为下标反转的元素,i<j保证只交换一次
        int k = len >> 1;
        while(j >= k){
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}

//做FFT,len必须为2的幂,on=1是DFT,on=-1是IDTF
void fft(Complex y[], int len, int on){
    change(y, len);//调用反转置换
    for(int h = 2; h <= len; h <<= 1){
        Complex wn(cos(-on * 2 * PI / h), sin(-on * 2 * PI / h));
        for(int j = 0; j < len; j += h){
            Complex w(1, 0);//初始化螺旋因子
            for(int k = j; k < j + h / 2; k++){//配对
                Complex u = y[k];
                Complex t = w * y[k + h / 2];
                y[k] = u + t;
                y[k + h / 2] = u - t;
                w = w * wn;//更新螺旋因子
            }
        }
    }
    if(on == -1){
        for(int i = 0; i < len; i++){
            y[i].x /= len;//IDTF
        }
    }
}

const int MAXN = 2e5 + 10;
Complex x1[MAXN], x2[MAXN];
char str1[MAXN >> 1], str2[MAXN >> 1];
int sum[MAXN];

int main(void){
    while(~scanf("%s%s", str1, str2)){
        int len1 = strlen(str1);
        int len2 = strlen(str2);
        int len = 1;
        while(len < len1 * 2 || len < len2 * 2) len <<= 1;
        //将str1,str2构造成两个多项式
        for(int i = 0; i < len1; i++){//倒存
            x1[i] = Complex(str1[len1 - i - 1] - '0', 0);
        }
        for(int i = len1; i < len; i++){
            x1[i] = Complex(0, 0);//不够的补0
        }
        for(int i = 0; i < len2; i++){
            x2[i] = Complex(str2[len2 - i - 1] - '0', 0);
        }
        for(int i = len2; i < len; i++){
            x2[i] = Complex(0, 0);
        }
        fft(x1, len, 1); //DFT(str1)
        fft(x2, len, 1); //DFT(str2);
        for(int i = 0; i < len; i++){
            x1[i] = x1[i] * x2[i];//点乘结果存入x1
        }
        fft(x1, len, -1);//IDFT(a*b)
        for(int i = 0; i < len; i++){
            sum[i] = (int)(x1[i].x + 0.5);//四舍五入
        }
        for(int i = 0; i < len; i++){
            sum[i + 1] += sum[i] / 10;
            sum[i] %= 10;
        }
        len = len1 + len2 - 1;//长度为len1和len2的两个数的乘积长度最大不超过len1+len2+1
        while(sum[len] <= 0 && len > 0) len--;//去前导0
        for(int i = len; i >= 0; i--){
            printf("%d", sum[i]);
        }
        puts("");
    }
    return 0;
}