题目大意：

求两个大整数的乘积, 两个大整数长度都不超过50000, 多组数据, 时限1s

大致思路:

作为FFT算法的一个开头的题, 恩还是仔细写了一下

这个题就是把整数视作是两个多项式, 每一位就是一项, 那么就相当于是两个最高次数不超过50000的多项式乘积之后在x = 10出的值, 那么这样就很简单了, 直接处理出其多项式然后用FFT计算即可, 试水题, 改进模板中…

FFT总结传送门：戳我

代码如下：

Result  :  Accepted     Memory  :  8380 KB     Time  :  140 ms

/*
 * Author: Gatevin
 * Created Time:  2015/7/14 16:07:01
 * File Name: HDU1402.cpp
 */
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

const double PI = acos(-1.0);

struct Complex
{
    double real, image;
    Complex(double _real, 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[140000];
void FFT(Complex* a, int len, int DFT)//对a进行DFT或者逆DFT, 结果存在a当中
{
    //Complex* A = new Complex[len]; 这么写会爆栈
    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;
}

char numA[50010], numB[50010];//以每一位为系数, 那么多项式长度不超过50000
Complex a[140000], b[140000];//对应的乘积的长度不会超过100000, 也就是不超过(1 << 17) = 131072
int ans[140000];
int main()
{
    while(~scanf("%s", numA))
    {
        int lenA = strlen(numA);
        int sa = 0;
        while((1 << sa) < lenA) sa++;
        scanf("%s", numB);
        int lenB = strlen(numB);
        int sb = 0;
        while((1 << sb) < lenB) sb++;
        //那么乘积多项式的次数不会超过(1 << (max(sa, sb) + 1))
        int len = (1 << (max(sa, sb) + 1));
        for(int i = 0; i < len; i++)
        {
            if(i < lenA) a[i] = Complex(numA[lenA - i - 1] - '0', 0);
            else a[i] = Complex(0, 0);
            if(i < lenB) b[i] = Complex(numB[lenB - i - 1] - '0', 0);
            else b[i] = Complex(0, 0);
        }
        FFT(a, len, 1);
        FFT(b, len, 1);//把A和B换成点值表达
        for(int i = 0; i < len; i++)//做点值表达的成乘法
            a[i] = a[i]*b[i];
        FFT(a, len, -1);//逆DFT换回原来的系数, 虚部一定是0
        for(int i = 0; i < len; i++)
            ans[i] = (int)(a[i].real + 0.5);//取整误差的处理
        for(int i = 0; i < len - 1; i++)//进位问题
        {
            ans[i + 1] += ans[i] / 10;
            ans[i] %= 10;
        }
        bool flag = 0;
        for(int i = len - 1; i >= 0; i--)//注意输出格式的调整即可
        {
            if(ans[i]) printf("%d", ans[i]), flag = 1;
            else if(flag || i == 0) printf("0");
        }
        putchar('\n');
    }
    return 0;
}