HDU 1402 A * B Problem Plus

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

 

Recommend

DOOM III   |   We have carefully selected several similar problems for you:  1695 1066 1042 1408 1405 

~~~~~~~~~~~~~~~~~~~~~~~~~~~

FFT~

需要注意的是这样多组数据a,b,c都要清零！

#include<cstdio>#include<cstring>#include<complex>#include<cmath>using namespace std;#define E complex<double>#define pi acos(-1)int n,m,l,r[200001],c[200001];E a[200001],b[200001];char ch[200001];void fft(E *u,int v){for(int i=0;i<n;i++) if(r[i]>i) swap(u[r[i]],u[i]);for(int i=1;i<n;i<<=1){E wn(cos(pi/i),v*sin(pi/i));for(int j=0;j<n;j+=(i<<1)){E w(1,0);for(int k=0;k<i;k++,w=w*wn){E x=u[k+j],y=u[i+j+k]*w;u[k+j]=x+y;u[i+j+k]=x-y;}}}if(v==-1) for(int i=0;i<n;i++) u[i]/=n;}int main(){while(scanf("%s",ch+1)==1){m=strlen(ch+1);l=0;for(int i=m-1;~i;i--) a[i]=(double)ch[m-i]-'0';scanf("%s",ch+1);n=strlen(ch+1);for(int i=n-1;~i;i--) b[i]=(double)ch[n-i]-'0';m+=n;for(n=1;n<=m;n<<=1) l++;for(int i=0;i<n;i++) r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));fft(a,1);fft(b,1);for(int i=0;i<n;i++) a[i]=a[i]*b[i];fft(a,-1);for(int i=0;i<n;i++) c[i]=(int)(a[i].real()+0.5);for(int i=0;i<=m;i++){c[i+1]+=c[i]/10;c[i]%=10;if(i==m && c[i+1]) m++;}while(!c[m] && m) m--;for(int i=m;~i;i--) printf("%d",c[i]);puts("");memset(c,0,sizeof(c));memset(a,0,sizeof(a));memset(b,0,sizeof(b));}return 0;}