問題來源：

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

題目描述：

Calculate A * B. 

Input

Each line will contain two integers A and B. Process toend 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

計算a*b，6位一乘，否則tle，而且只能交G++，另外傅立葉演算法也可以

程式程式碼：

#include<stdio.h>
#include<string.h>
#define N 50100
int a[N],b[N];
long long A[N],B[N],ans[N];
char s1[N],s2[N];
int main()
{
    int i,j,len1,len2,l,r,L,R,x;
    long long t,p,c;
    while(scanf("%s%s",s1,s2)!=EOF)
    {
     if(s1[0]=='0'||s2[0]=='0')
     {
     printf("0\n");
     continue;
     }
        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));
         
        len1=strlen(s1);
        len2=strlen(s2);
        l=len1-1;
        r=len2-1;
        for(i=0;i<len1;i++)                     
        a[l--]=(s1[i]-'0');
        for(j=0;j<len2;j++)
        b[r--]=(s2[j]-'0');
        L=0;R=0;
    
         for(i=0;i<len1;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];
                                              
        for(i=0; i<len2; 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];
  
        for(i=0;i<L;i++)
        {
            c=0;
            for(j=0; j<R; j++)
            {
                t=A[i]*B[j]+ans[i+j]+c;            
                ans[i+j]=t%1000000;
                c=t/1000000;
               
            }
            if(c)
                ans[i+j]=c;
        }
          
      
        for(x=L+R;ans[x]==0;x--){
        }
        printf("%lld",ans[x]);
        
        for(x--; x>=0;x--)
        printf("%06lld",ans[x]);                 
        printf("\n");
    }
    return 0;
}