HDU 1402 A * B Problem Plus（FFT 大整数乘法）

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

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18633 Accepted Submission(s): 4225

Problem Description
Calculate A * B.

Input
Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

Output
For each case, output A * B in one line.

Sample Input
1
2
1000
2

Sample Output
2
2000

【思路分析】FFT做大整数乘法。
【AC代码】

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define LL long long

const double eps(1e-8);

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FFT或者逆DFT,结果存在a中
{
    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 str1[50005],str2[50005];//以每一位数字做为多项式的系数
Complex a[140000],b[140000];//乘积的长度不会超过100000
int ans[140000];
int main()
{
    while(~scanf("%s",str1))
    {
        int len1=strlen(str1);
        int sa=0;
        while((1<<sa)<len1)sa++;
        scanf("%s",str2);
        int len2=strlen(str2);
        int sb=0;
        while((1<<sb)<len2)sb++;
        int len=(1<<(max(sa,sb)+1));//成积多项式的次数最多不会超过max(sa,sb)+1
        for(int i=0; i<len; i++)
        {
            if(i<len1)
            {
                a[i]=Complex(str1[len1-i-1]-'0',0);
            }
            else
            {
                a[i]=Complex(0,0);
            }
            if(i<len2)
            {
                b[i]=Complex(str2[len2-i-1]-'0',0);
            }
            else
            {
                b[i]=Complex(0,0);
            }
        }
        FFT(a,len,1);//把a,b都换成点表达式
        FFT(b,len,1);
        for(int i=0; i<len; i++)//做点表达式的乘法
        {
            a[i]=a[i]*b[i];
        }
        FFT(a,len,-1);//逆DFT换回原来的次数 ，虚部一定都是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--)//输出格式的调整
        {
            if(ans[i])
                printf("%d",ans[i]),flag=1;
            else if(flag||i==0)
            {
                printf("0");
            }
        }
        printf("\n");
    }
    return 0;
}