大数模板

//高精度模板，注意本模板并没有考虑中间会出现负数的情况    

#include <stdio.h>    

#include <string.h>    

#include <stdlib.h>    

#include <math.h>    

#include <assert.h>    

#include <ctype.h>    

#include <map>    

#include <string>    

#include <set>    

#include <bitset>    

#include <utility>    

#include <algorithm>    

#include <vector>    

#include <stack>    

#include <queue>    

#include <iostream>    

#include <fstream>    

#include <list>      

     

using  namespace  std;      

     

const  int MAXL = 500;      

struct  BigNum      

{      

    int  num[MAXL];      

    int  len;      

};      

     

//高精度比较 a > b return 1,a == b return 0; a < b return -1;      

int  Comp(BigNum &a,BigNum &b)      

{      

    int  i;      

    if(a.len != b.len) return (a.len > b.len) ? 1 : -1;      

    for(i = a.len-1; i >= 0; i--)      

        if(a.num[i] != b.num[i]) return  (a.num[i] > b.num[i]) ? 1 : -1;      

    return  0;      

}      

     

//高精度加法      

BigNum  Add(BigNum &a,BigNum &b)      

{      

    BigNum c;      

    int  i,len;      

    len = (a.len > b.len) ? a.len : b.len;      

    memset(c.num,sizeof(c.num));      

    for(i = 0; i < len; i++)      

    {      

        c.num[i] += (a.num[i]+b.num[i]);      

        if(c.num[i] >= 10)      

        {      

            c.num[i+1]++;      

            c.num[i] -= 10;      

        }      

    }      

    if(c.num[len]) len++;      

    c.len = len;      

    return  c;      

}      

//高精度减法,保证a >= b      

BigNum Sub(BigNum &a,BigNum &b)      

{      

    BigNum  c;      

    int  i,sizeof(c.num));      

    for(i = 0; i < len; i++)      

    {      

        c.num[i] += (a.num[i]-b.num[i]);      

        if(c.num[i] < 0)      

        {      

            c.num[i] += 10;      

            c.num[i+1]--;      

        }      

    }      

    while(c.num[len] == 0 && len > 1) len--;      

    c.len = len;      

    return  c;      

}      

//高精度乘以低精度，当b很大时可能会发生溢出int范围,具体情况具体分析      

//如果b很大可以考虑把b看成高精度      

BigNum Mul1(BigNum &a,int  &b)      

{      

    BigNum c;      

    int  i,len;      

    len = a.len;      

    memset(c.num,sizeof(c.num));      

    //乘以0，直接返回0      

    if(b == 0)       

    {      

        c.len = 1;      

        return  c;      

    }      

    for(i = 0; i < len; i++)      

    {      

        c.num[i] += (a.num[i]*b);      

        if(c.num[i] >= 10)      

        {      

            c.num[i+1] = c.num[i]/10;      

            c.num[i] %= 10;      

        }      

    }      

    while(c.num[len] > 0)      

    {      

        c.num[len+1] = c.num[len]/10;      

        c.num[len++] %= 10;      

    }      

    c.len = len;       

    return  c;      

}      

     

//高精度乘以高精度,注意要及时进位，否则肯能会引起溢出,但这样会增加算法的复杂度，      

//如果确定不会发生溢出,可以将里面的while改成if      

BigNum  Mul2(BigNum &a,BigNum &b)      

{      

    int     i,j,len = 0;      

    BigNum  c;      

    memset(c.num,sizeof(c.num));      

    for(i = 0; i < a.len; i++)      

        for(j = 0; j < b.len; j++)      

        {      

            c.num[i+j] += (a.num[i]*b.num[j]);      

            if(c.num[i+j] >= 10)      

            {      

                c.num[i+j+1] += c.num[i+j]/10;      

                c.num[i+j] %= 10;      

            }      

        }      

    len = a.len+b.len-1;      

    while(c.num[len-1] == 0 && len > 1) len--;      

    if(c.num[len]) len++;      

    c.len = len;      

    return  c;      

}      

     

//高精度除以低精度,除的结果为c,余数为f      

void Div1(BigNum &a,int &b,BigNum &c,int &f)      

{      

    int  i,len = a.len;      

    memset(c.num,sizeof(c.num));      

    f = 0;      

    for(i = a.len-1; i >= 0; i--)      

    {      

        f = f*10+a.num[i];      

        c.num[i] = f/b;      

        f %= b;      

    }      

    while(len > 1 && c.num[len-1] == 0) len--;      

    c.len = len;      

}      

//高精度*10      

void  Mul10(BigNum &a)      

{      

    int  i,len = a.len;      

    for(i = len; i >= 1; i--)      

        a.num[i] = a.num[i-1];      

    a.num[i] = 0;      

    len++;      

    //if a == 0      

    while(len > 1 && a.num[len-1] == 0)  len--;      

}      

     

//高精度除以高精度，除的结果为c，余数为f      

void Div2(BigNum &a,BigNum &b,BigNum &f)      

{      

    int  i,sizeof(c.num));      

    memset(f.num,sizeof(f.num));      

    f.len = 1;      

    for(i = len-1;i >= 0;i--)      

    {      

        Mul10(f);      

        //余数每次乘10      

        f.num[0] = a.num[i];      

        //然后余数加上下一位      

        ///利用减法替换除法      

        while(Comp(f,b) >= 0)      

        {      

            f = Sub(f,b);      

            c.num[i]++;      

        }      

    }      

    while(len > 1 && c.num[len-1] == 0)len--;      

    c.len = len;      

}   

void  print(BigNum &a)      

{      

    int  i;      

    for(i = a.len-1; i >= 0; i--)      

        printf("%d",a.num[i]);      

    puts("");      

}      

//将字符串转为大数存在BigNum结构体里面      

BigNum ToNum(char *s)      

{      

    int i,j;      

    BigNum  a;      

    a.len = strlen(s);      

    for(i = 0,j = a.len-1; s[i] != ''; i++,j--)      

        a.num[i] = s[j]-'0';      

    return  a;      

}      

     

void  Init(BigNum &a,char *s,int &tag)   

{   

    int  i = 0,j = strlen(s);   

    if(s[0] == '-') {j--; i++; tag *= -1;}   

    a.len = j;   

    for(; s[i] != ''; i++,j--)   

        a.num[j-1] = s[i]-'0';   

    //print(a);   

}   

  

int  main()      

{      

    BigNum a,b;   

    char  s1[100],s2[100];   

    while(scanf("%s %s",s1,s2) != EOF)   

    {   

        int tag = 1;   

        Init(a,tag);   

        Init(b,s2,tag);   

        a = Mul2(a,b);   

        if(a.len == 1 && a.num[0] == 0)   

        {   

            puts("0");   

        }   

        else    

        {   

            if(tag < 0) putchar('-');   

            print(a);   

        }   

    }   

    return 0;   

}