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 = ;
 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=,y=;}
     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=,j=; i<n; ++i) {
         if (i < j) swap(a[i],a[j]);
         for (int k=n>>; (j^=k)<k; k>>=); //妙啊！！！
     }
     for (int m=; m<=n; m<<=) {
         Complex w1 = Complex(cos(*pi/m),ty*sin(*pi/m));
         for (int i=; i<n; i+=m) {
             Complex w = Complex(,);
             for (int k=; k<(m>>); ++k) {
                 Complex t = w * a[i+k+(m>>)];
                 a[i+k+(m>>)] = 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 = ;
         while (n < (len1+len2)) n <<= ;
         for (int i=; i<n; ++i) {
             if (i < len1) A[i] = Complex(a[len1-i-]-'',);
             else A[i] = Complex(,);
             if (i < len2) B[i] = Complex(b[len2-i-]-'',);
             else B[i] = Complex(,);
         }
         FFT(A,n,);
         FFT(B,n,);
         for (int i=; i<n; ++i) A[i] = A[i] * B[i];
         FFT(A,n,-);
         for (int i=; i<n; ++i)
             ans[i] = (int)(A[i].x/n+0.5);
         for (int i=; i<n-; ++i) {
             ans[i+] += (ans[i]/);
             ans[i] %= ;
         }
         bool fir = false;
         for (int i=n-; i>=; --i) {
             if (ans[i]) printf("%d",ans[i]),fir = true;
             else if (fir || i==) printf("");
         }
         puts("");
     }
     return ;
 }