高精度板子

#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, 0, 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, len;        
    len = (a.len > b.len) ? a.len : b.len;        
    memset(c.num, 0, 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, 0, 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, 0, 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, 0, 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 &c, BigNum &f)        
{        
    int  i, len = a.len;        
    memset(c.num, 0, sizeof(c.num));        
    memset(f.num, 0, 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)   //输出大数     
{        
    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)   //将字符串转化为大数  
{     
    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';  
}     
    
int main(void)        
{        
    BigNum a, b;     
    char  s1[100], s2[100];     
    while(scanf("%s %s", s1, s2) != EOF)     
    {     
        int tag = 1;     
        Init(a, s1, tag);    //将字符串转化为大数  
        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;     
}