HDU1042 A * B Problem Plus

 

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24620    Accepted Submission(s): 6271

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

 

 

分析

一直不过，最后发现数组开小了qwq。

思路：把每个数分解成多项式

$n = a_0*10^0+a_1*10^1+...a_k*10^k$

然后多项式乘法，FFT模板题。

code 

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<iostream>

using namespace std;

const int N = 200100;
const double pi = acos(-1.0);
char a[N],b[N];
int ans[N]; //数组大小！

struct Complex{
    double x,y;
    Complex() {x=0,y=0;}
    Complex(double _x,double _y) {x = _x,y = _y;}
}A[N],B[N];
Complex operator + (Complex a,Complex b) {
    return Complex(a.x+b.x,a.y+b.y);
}
Complex operator - (Complex a,Complex b) {
    return Complex(a.x-b.x,a.y-b.y);
}
Complex operator * (Complex a,Complex b) {
    return Complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
}
void FFT(Complex *a,int n,int ty) {
    for (int i=0,j=0; i<n; ++i) {
        if (i < j) swap(a[i],a[j]);
        for (int k=n>>1; (j^=k)<k; k>>=1); //妙啊！！！
    }
    for (int m=2; m<=n; m<<=1) {
        Complex w1 = Complex(cos(2*pi/m),ty*sin(2*pi/m));
        for (int i=0; i<n; i+=m) {
            Complex w = Complex(1,0);
            for (int k=0; k<(m>>1); ++k) {
                Complex t = w * a[i+k+(m>>1)];
                a[i+k+(m>>1)] = a[i+k] - t;
                a[i+k] = a[i+k] + t;
                w = w * w1;
            }
        }
    }
}
int main () {
    while (scanf("%s%s",a,b)!=EOF) {
        int len1 = strlen(a),len2 = strlen(b);
        int n = 1;
        while (n < (len1+len2)) n <<= 1;
        for (int i=0; i<n; ++i) {
            if (i < len1) A[i] = Complex(a[len1-i-1]-'0',0);
            else A[i] = Complex(0,0);
            if (i < len2) B[i] = Complex(b[len2-i-1]-'0',0);
            else B[i] = Complex(0,0);
        }
        FFT(A,n,1);
        FFT(B,n,1);
        for (int i=0; i<n; ++i) A[i] = A[i] * B[i];
        FFT(A,n,-1);
        for (int i=0; i<n; ++i) 
            ans[i] = (int)(A[i].x/n+0.5);
        for (int i=0; i<n-1; ++i) {
            ans[i+1] += (ans[i]/10);
            ans[i] %= 10;
        }
        bool fir = false;
        for (int i=n-1; i>=0; --i) {
            if (ans[i]) printf("%d",ans[i]),fir = true;
            else if (fir || i==0) printf("0");
        }
        puts("");
    }
    return 0;
}