HDU 1402 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): 11884    Accepted Submission(s): 2052

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

1210002

 

Sample Output

22000

 

Author

DOOM III

大数相乘的 FFT模板，来自kuangbin大神

  

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define PI acos(-1.0)using namespace std;struct complex{double r,i;complex (double _r=0.0,double _i=0.0){r=_r;i=_i;}complex operator +(const complex &b){return complex(r+b.r,i+b.i);}complex operator -(const complex &b){return complex(r-b.r,i-b.i);}complex operator *(const complex &b){return complex(r*b.r-i*b.i,r*b.i+i*b.r);}};void change(complex y[],int len){int i,j,k;for(i=1,j=len/2;i<len-1;i++){if(i<j)swap(y[i],y[j]);k=len/2;while(j>=k){j-=k;k/=2;}if(j<k)j+=k;}}void fft(complex y[],int len,int on){change(y,len);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);for(int k=j;k<j+h/2;k++){complex u=y[k];complex t=w*y[k+h/2];y[k]=u+t;y[k+h/2]=u-t;w=w*wn;}}}if(on==-1)for(int i=0;i<len;i++)y[i].r/=len;}const int Maxn=200010;complex x1[Maxn],x2[Maxn];char str1[Maxn/2],str2[Maxn/2];int sum[Maxn];int main(){freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);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;for(int i=0;i<len1;i++)x1[i]=complex(str1[len1-i-1]-'0',0);for(int i=len1;i<len;i++)x1[i]=complex(0,0);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);fft(x2,len,1);for(int i=0;i<len;i++)x1[i]=x1[i]*x2[i];fft(x1,len,-1);for(int i=0;i<len;i++)sum[i]=int(x1[i].r+0.5);for(int i=0;i<len;i++){sum[i+1]+=sum[i]/10;sum[i]%=10;}len=len1+len2-1;while(sum[len]<=0&&len>0)len--;for(int i=len;i>=0;i--)printf("%c",sum[i]+'0');printf("\n");}return 0;}