hdu 1402 A * B Problem Plus

A * B Problem Plus 

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

求A*B，多存几位就可以过，或者用FFT

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int a[50100];
long long A[50100];
long long B[50100];
int b[50100];
long long ans[1000001];
int main()
{
    string s1,s2;
    while(cin>>s1>>s2)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(A,0,sizeof(A));
        memset(B,0,sizeof(B));
        memset(ans,0,sizeof(ans));
        int l=0,r=0;
        if(s1=="0"||s2=="0")
        {
            printf("0
");
            continue;
        }
        for(int i=s1.size()-1; i>=0; i--)
            a[l++]=s1[i]-'0';
        for(int i=s2.size()-1; i>=0; i--)
            b[r++]=s2[i]-'0';
        int L=0,R=0;
        for(int i=0; i<l; i+=6)
        {
            A[L++]=a[i+5]*100000+a[i+4]*10000+a[i+3]*1000+a[i+2]*100+a[i+1]*10+a[i];
            //cout<<A[L-1];
        }
        for(int i=0; i<r; i+=6)
        {
            B[R++]=b[i+5]*100000+b[i+4]*10000+b[i+3]*1000+b[i+2]*100+b[i+1]*10+b[i];
            //cout<<B[R-1];
        }
        long long p;
        for(int i=0; i<L; i++)
        {
            long long c=0;
            for(int j=0; j<R; j++)
            {
                long long temp=A[i]*B[j]+ans[i+j]+c;
                //cout<<temp<<endl;
                ans[i+j]=temp%1000000;
                c=temp/1000000;
                p=j;
            }
            if(c)
                ans[i+p+1]=c;
        }
        int x=L+R;
        for(x; ans[x]==0; x--);
        printf("%lld",ans[x]);
        x--;
        for(x; x>=0; x--)
            printf("%06lld",ans[x]);
        cout<<endl;
    }
}

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
const double PI=acos(-1.0);
struct Complex
{
    double x,y;
    Complex(double _x=0.0,double _y=0.0)
    {
        x=_x;
        y=_y;
    }
    Complex operator-(const Complex &b)const{return Complex(x-b.x,y-b.y);}
    Complex operator+(const Complex &b)const{return Complex(x+b.x,y+b.y);}
    Complex operator*(const Complex &b)const{return Complex(x*b.x-y*b.y,x*b.y+y*b.x);}
};
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].x/=len;
        }
    }
}
const int MAXN=200010;
Complex x1[MAXN],x2[MAXN];
char str1[MAXN/2],str2[MAXN/2];
int sum[MAXN];
int main()
{
    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[len1-1-i]-'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[len2-1-i]-'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].x+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("
");
    }
    return 0;
}