HDU 1402(A * B Problem Plus-FFT速度测试)

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15111    Accepted Submission(s): 2942

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
   
   
   
   

 

Author

DOOM III

 

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#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair 
#define eps (1e-1)
#define MAXN (200000+10)
#define pi ((double)3.1415926535897932384626)
typedef long long ll;
typedef complex<double> cd;

class fft { 
public:
	cd A[MAXN];
	int n,l;
	void brc(cd *A,int l) {
		int i,j,k;
		for(i=1,j=l>>1;i<l-1;i++) {
			if (i<j) swap(A[i],A[j]);
			k=l>>1;
			while(j>=k) {
				j-=k;
				k>>=1;
			} 
			j+=k;
		} 	
	}	
	void DFT(int l,int on) //on 	
	{
		brc(A,l);
		
		for(int h=2;h<=l;h<<=1) {
			cd wn=cd(cos(on*2*pi/h),sin(on*2*pi/h));
			for(int j=0;j<l;j+=h) {
				cd w=cd(1,0);
				
				for(int k=j;k<j+h/2;k++) {
					cd u=A[k],t=w*A[k+h/2];
					A[k]=u+t;
					A[k+h/2]=u-t;
					w*=wn;
				}
			}
			
		}
		if (on==-1) Rep(i,l) A[i]/=l;
		
		//DFT = 逆矩阵=-A/l  
		
	}
	void mem(int _n) {
		MEM(A)  n=_n;
		l=1; 
		while(l<n) l<<=1; 
		l<<=1;
	}
	
	void scan(char *a,int n) {
		MEM(A)
		Rep(i,n) {
			A[i]=cd(a[n-i-1]-'0',0);
		}		
	}
	
}S1,S2;
char s1[MAXN],s2[MAXN];
int ans[MAXN];

int main()
{
//	freopen("hdu1402.in","r",stdin);
//	freopen(".out","w",stdout);
	
	while(scanf("%s%s",s1,s2)==2) {
		int n=strlen(s1),m=strlen(s2);
		int N=max(n,m);
		S1.mem(N),S2.mem(N);
		
		S1.scan(s1,n);
		S2.scan(s2,m);
		S1.DFT(S1.l,1);
		S2.DFT(S2.l,1);
		Rep(i,S1.l) S1.A[i]*=S2.A[i];
		S1.DFT(S1.l,-1);
		
		MEM(ans)
		Rep(i,S1.l) 
		{
			ans[i]+=(int)(S1.A[i].real()+0.1);
			ans[i+1]+=ans[i]/10;
			ans[i]%=10;
		} 
		
		int L=S1.l;
		while(L&&!ans[L]) --L;
		RepD (i,L) printf("%d",ans[i]);
		
		putchar('\n');
	}
	
	
	return 0;
}