大数(高精度数)模板(分享)
大数(高精度数)模板(分享)
发布时间:2016-12-28 来源:查字典编辑
摘要:复制代码代码如下:#include#include#include#include#include#include#include#incl...

复制代码 代码如下:

#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;

}

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