hdu1042(好大的数啊！)

http://acm.hdu.edu.cn/showproblem.php?pid=1402

我的代码，由于时间的问题，以后找机会优化了！（超时）

#include"iostream"
#define M 100010
using namespace std;
char ch1[M],ch2[M];
int a[M],b[M];
int c[M],d[M];
int num[M];
int H[M];
int t,k,g,v;
int i,j;
int La,Lb;
int sign=0;
void Add(int x  ,int y)
{
     sign=0;     t=0;
     int p,q,flag=0;
  if(x>y)
  {
    for(p=0,q=0;p<y,q<y; p++,q++)
    {
          num[t++]=(d[p]+H[q]+flag)%10;
        flag=(d[p]+H[q]+flag)%10;
    }
    for(p=y; p<x ; p++)
    {
        num[t++]=(d[p]+flag)%10;
        flag=(d[p]+flag)/10;
    }
  }
  else if(x==y)
  {
    for(p=0,q=0;p<x,q<y; p++,q++)
    {
          num[t++]=(d[p]+H[q]+flag)%10;
        flag=(d[p]+H[q]+flag)/10;
    }
  }

  else
  {
      flag=0;
    for(p=0,q=0;p<x,q<x; p++,q++)
    {
          num[t++]=(H[p]+d[q]+flag)%10;
        flag=(H[p]+d[q]+flag)/10;
    }
    for(q=x;q<y;q++)
    {
        num[t++]=(H[q]+flag)%10;
        flag=(H[q]+flag)/10;
    }
  }

  while(flag)
  {
    num[t++]=flag%10;
    flag/=10;
  }
 
  for(int yy=0; yy<t;yy++)
  {
      d[yy]=num[yy];
      if(d[yy]==0) sign++;    //sign用于标记都为结果0的情况
  }
     g=t;
     memset(H , 0 , sizeof(H));
     memset(num, 0, sizeof(num));
}

void BigN()
{
    int flag=0;
    int mark=0;
   if(La>Lb)
   {
       for(i=Lb-1;i>=0;i--)
       {
           k=0;
           for(j=La-1;j>=0;j--)
           {
               c[k++]=(b[i]*a[j]+flag)%10;
               flag=(b[i]*a[j]+flag)/10;
           }

            while(flag)
            {
              c[k++]=flag%10;
              flag/=10;
            }
            
            //末尾赋值0
             v=0;
            for(int xx=0; xx<mark; xx++)
                H[v++]=0;
            for(int m=0;m<k;m++) 
                H[v++]=c[m];

            memset(c,0,sizeof(c));
            //****************************  
             Add(g , v);
             mark++;
             
       }
   }
   else 
   {
        for(i=La-1;i>=0;i--)
       {
           k=0;
           for(j=Lb-1;j>=0;j--)
           {
               c[k++]=(a[i]*b[j]+flag)%10;
               flag=(a[i]*b[j]+flag)/10;
           }
            while(flag)
            {
             c[k++]=flag%10;
              flag/=10;
            }

            //末尾赋值0
             v=0;
            for(int xx=0; xx<mark; xx++)
                H[v++]=0;
            for(int m=0;m<k;m++) 
                H[v++]=c[m];
            //**************************** 
             Add(g , v);

               mark++;
       } 
   }
  
}

int main()
{
     while(cin>>ch1>>ch2)
     {
         t=0;k=0;g=0;

         La=strlen(ch1);
         Lb=strlen(ch2);

         for(i=0;i<La;i++) a[i]=ch1[i]-'0';
           for(i=0;i<Lb;i++) b[i]=ch2[i]-'0';

           BigN();

           if(sign!=t)
           {
              for(i=t-1;i>=0;i--)
              cout<<d[i];
               cout<<endl;
           }
           else cout<<"0"<<endl;
     }
     return 0;
}

一位dn的，神人啊！

http://www.cppblog.com/misschuer/archive/2010/01/16/80356.html

#include<iostream>
#include<cmath>
using namespace std;
typedef struct vir{
        double re,im;
        vir(){}
        vir(double a,double b){re=a;im=b;}
        vir operator +(const vir &b){ return vir(re+b.re,im+b.im);}
        vir operator -(const vir &b){ return vir(re-b.re,im-b.im);}
        vir operator *(const vir &b){ return vir(re*b.re-im*b.im,re*b.im+b.re*im);}
}vir;
vir x1[200005],x2[200005];
const double Pi = acos(-1.0);
void change(vir *x,int len,int loglen)
{
        int i,j,k,t;
        for(i=0;i<len;i++)
        {
                t = i;
                for(j=k=0;j<loglen;j++,t>>=1)
                        k = (k<<1)|(t&1);
                if(k<i)
                {
                        vir wt =  x[k];
                        x[k] = x[i];
                        x[i] = wt;
                }
        }
}
void fft(vir *x,int len,int loglen)
{
        int i,j,t,s,e;
        change(x,len,loglen);
        t = 1;
        for(i=0;i<loglen;i++,t<<=1)
        {
                s = 0;
                e = s + t;
                while(s<len)
                {
                        vir a,b,wo(cos(Pi/t),sin(Pi/t)),wn(1,0);
                        for(j=s;j<s+t;j++)
                        {
                                a = x[j];
                                b = x[j+t]*wn;
                                x[j] = a + b;
                                x[j+t] = a - b;
                                wn =wn*wo;
                        }
                        s = e+t;
                        e = s+t;
                }
        }
}

void dit_fft(vir *x,int len,int loglen)
{
        int i,j,s,e,t=1<<loglen;
        for(i=0;i<loglen;i++)
        {
                t>>=1;
                s=0;
                e=s+t;
                while(s<len)
                {
                        vir a,b,wn(1,0),wo(cos(Pi/t),-sin(Pi/t));
                        for(j=s;j<s+t;j++)
                        {
                                a = x[j]+x[j+t];
                                b = (x[j]-x[j+t])*wn;
                                x[j] = a;
                                x[j+t] = b;
                                wn = wn*wo;
                        }
                        s = e+t;
                        e = s+t;
                }
        }
        change(x,len,loglen);
        for(i=0;i<len;i++)
                x[i].re/=len;
}


int main()
{
        char a[100005],b[100005];
        int i,len1,len2,t,over,len,loglen;
        
        while(scanf("%s%s",a,b)!=EOF)
        {
                len1 = strlen(a)<<1;
                len2 = strlen(b)<<1;
                len = 1;
                loglen = 0;
                while(len<len1)
                {
                        len<<=1;
                        loglen++;
                }
                while(len<len2)
                {
                        len<<=1;
                        loglen++;
                }
                for(i=0;a[i]!='\0';i++)
                {
                        x1[i].re = a[i]-'0';
                        x1[i].im = 0;
                }
                for(;i<len;i++)
                        x1[i].re = x1[i].im = 0;
                for(i=0;b[i]!='\0';i++)
                {
                        x2[i].re = b[i]-'0';
                        x2[i].im = 0;
                }
                for(;i<len;i++)
                        x2[i].re = x2[i].im = 0;
                fft(x1,len,loglen);
                fft(x2,len,loglen);
                for(i=0;i<len;i++)
                        x1[i] = x1[i]*x2[i];
                dit_fft(x1,len,loglen);
                for(i=(len1+len2)/2-2,over=loglen=0;i>=0;i--)
                {
                        t = x1[i].re + over + 0.5;
                        a[loglen++] = t%10;
                        over =  t/10;
                }
                while(over)
                {
                        a[loglen++] = over%10;
                        over /= 10;
                }
                for(loglen--;loglen>=0&&!a[loglen];loglen--);
                if(loglen<0)
                        putchar('0');
                else
                        for(;loglen>=0;loglen--)
                                putchar(a[loglen]+'0');
                putchar('\n');
        }
        return 0;
}