大数运算和大数类

1、首先看一个小问题：

整型数组、字符型数组、vector在未初始化的情况下：

int main()
{
int data[5];
char c[5];
vector<int> v(5);
vector<char> v1(5);
vector<int> v2;

cout << "int[]:	";
for (int i = 0; i < 5; i++)
{
cout << data[i] << " ";
}
cout << endl << endl;

cout << "char[]:	";
for (int i = 0; i < 5; i++)
{
cout << c[i] << ",";
}
cout << endl << endl;

cout << "vector(int):	";
for (int i : v)
{
cout << i << " ";
}
cout << endl << endl;

cout << "vector(char):	";
for (auto i : v1)
{
cout << i << ",";
}
cout << endl << endl;

cout << "vector:	";
for (int i : v2)
{
cout << i << ",";
}
cout << endl << endl;

return 0;
}

vs测试结果：

（1）整型数组在未初始化时，内存中是乱码；

（2）字符型数组在未初始化时，内存中是乱码；

（3）vector指定容量，int型时，初始化为0；

（4）vector指定容量，char型时，初始化为可打印的空白字符；

（5）vector不指定容量，默认0；

2、简单的大数相加：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;

#define N 500

//25369874522455633 + 68452156965
void bigNumAdd(string a, string b)
{
int carry = 0; //表示进位

//字符串反转(真正的数字位数 和 字符串下标 刚好是相反的，加法应该从数字低位算起) 且 字符串变数字
int m
= { 0 };
int n
= { 0 };
int len1 = a.length(), len2 = b.length();
for (int i = 0; i < len1; i++)
{
m[i] = a[len1 - i - 1] - '0';   //加法应该从数字低位算起
}
for (int i = 0; i < len2; i++)
{
n[i] = b[len2 - i - 1] - '0';   //加法应该从数字低位算起
}

//位运算————加法应该从数字低位算起
int len = len1>len2 ? len1 + 1 : len2 + 1;
string sum(len, 0);
int k;
for (k = 0; k < len1 || k < len2; k++)
{
sum[k] = (m[k] + n[k] + carry) % 10 + '0';
carry = (m[k] + n[k] + carry) / 10;
}

//处理最高位进位
if (carry)
{
sum[k] = carry + '0';
}

//结果字符串反转 并 输出
reverse(sum.begin(), sum.end());
cout << sum << endl;
}

int main()
{
string a, b;
cin >> a >> b;

bigNumAdd(a, b);

return 0;
}

3、大数类：

转自：http://blog.csdn.net/hackbuteer1/article/details/6595881

#include<iostream>
#include<string>
#include<iomanip>
#include<algorithm>
using namespace std;

#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4

class BigNum
{
private:
int a[500];    //可以控制大数的位数
int len;       //大数长度
public:
BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数
BigNum(const int);       //将一个int类型的变量转化为大数
BigNum(const char*);     //将一个字符串类型的变量转化为大数
BigNum(const BigNum &);  //拷贝构造函数
BigNum &operator=(const BigNum &);   //重载赋值运算符，大数之间进行赋值运算

friend istream& operator>>(istream&,  BigNum&);   //重载输入运算符
friend ostream& operator<<(ostream&,  BigNum&);   //重载输出运算符

BigNum operator+(const BigNum &) const;   //重载加法运算符，两个大数之间的相加运算
BigNum operator-(const BigNum &) const;   //重载减法运算符，两个大数之间的相减运算
BigNum operator*(const BigNum &) const;   //重载乘法运算符，两个大数之间的相乘运算
BigNum operator/(const int   &) const;    //重载除法运算符，大数对一个整数进行相除运算

BigNum operator^(const int  &) const;    //大数的n次方运算
int    operator%(const int  &) const;    //大数对一个int类型的变量进行取模运算
bool   operator>(const BigNum & T)const;   //大数和另一个大数的大小比较
bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较

void print();       //输出大数
};
BigNum::BigNum(const int b)     //将一个int类型的变量转化为大数
{
int c,d = b;
len = 0;
memset(a,0,sizeof(a));
while(d > MAXN)
{
c = d - (d / (MAXN + 1)) * (MAXN + 1);
d = d / (MAXN + 1);
a[len++] = c;
}
a[len++] = d;
}
BigNum::BigNum(const char*s)     //将一个字符串类型的变量转化为大数
{
int t,k,index,l,i;
memset(a,0,sizeof(a));
l=strlen(s);
len=l/DLEN;
if(l%DLEN)
len++;
index=0;
for(i=l-1;i>=0;i-=DLEN)
{
t=0;
k=i-DLEN+1;
if(k<0)
k=0;
for(int j=k;j<=i;j++)
t=t*10+s[j]-'0';
a[index++]=t;
}
}
BigNum::BigNum(const BigNum & T) : len(T.len)  //拷贝构造函数
{
int i;
memset(a,0,sizeof(a));
for(i = 0 ; i < len ; i++)
a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n)   //重载赋值运算符，大数之间进行赋值运算
{
int i;
len = n.len;
memset(a,0,sizeof(a));
for(i = 0 ; i < len ; i++)
a[i] = n.a[i];
return *this;
}
istream& operator>>(istream & in,  BigNum & b)   //重载输入运算符
{
char ch[MAXSIZE*4];
int i = -1;
in>>ch;
int l=strlen(ch);
int count=0,sum=0;
for(i=l-1;i>=0;)
{
sum = 0;
int t=1;
for(int j=0;j<4&&i>=0;j++,i--,t*=10)
{
sum+=(ch[i]-'0')*t;
}
b.a[count]=sum;
count++;
}
b.len =count++;
return in;

}
ostream& operator<<(ostream& out,  BigNum& b)   //重载输出运算符
{
int i;
cout << b.a[b.len - 1];
for(i = b.len - 2 ; i >= 0 ; i--)
{
cout.width(DLEN);
cout.fill('0');
cout << b.a[i];
}
return out;
}

BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算
{
BigNum t(*this);
int i,big;      //位数
big = T.len > len ? T.len : len;
for(i = 0 ; i < big ; i++)
{
t.a[i] +=T.a[i];
if(t.a[i] > MAXN)
{
t.a[i + 1]++;
t.a[i] -=MAXN+1;
}
}
if(t.a[big] != 0)
t.len = big + 1;
else
t.len = big;
return t;
}
BigNum BigNum::operator-(const BigNum & T) const   //两个大数之间的相减运算
{
int i,j,big;
bool flag;
BigNum t1,t2;
if(*this>T)
{
t1=*this;
t2=T;
flag=0;
}
else
{
t1=T;
t2=*this;
flag=1;
}
big=t1.len;
for(i = 0 ; i < big ; i++)
{
if(t1.a[i] < t2.a[i])
{
j = i + 1;
while(t1.a[j] == 0)
j++;
t1.a[j--]--;
while(j > i)
t1.a[j--] += MAXN;
t1.a[i] += MAXN + 1 - t2.a[i];
}
else
t1.a[i] -= t2.a[i];
}
t1.len = big;
while(t1.a[t1.len - 1] == 0 && t1.len > 1)
{
t1.len--;
big--;
}
if(flag)
t1.a[big-1]=0-t1.a[big-1];
return t1;
}

BigNum BigNum::operator*(const BigNum & T) const   //两个大数之间的相乘运算
{
BigNum ret;
int i,j,up;
int temp,temp1;
for(i = 0 ; i < len ; i++)
{
up = 0;
for(j = 0 ; j < T.len ; j++)
{
temp = a[i] * T.a[j] + ret.a[i + j] + up;
if(temp > MAXN)
{
temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
up = temp / (MAXN + 1);
ret.a[i + j] = temp1;
}
else
{
up = 0;
ret.a[i + j] = temp;
}
}
if(up != 0)
ret.a[i + j] = up;
}
ret.len = i + j;
while(ret.a[ret.len - 1] == 0 && ret.len > 1)
ret.len--;
return ret;
}
BigNum BigNum::operator/(const int & b) const   //大数对一个整数进行相除运算
{
BigNum ret;
int i,down = 0;
for(i = len - 1 ; i >= 0 ; i--)
{
ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
}
ret.len = len;
while(ret.a[ret.len - 1] == 0 && ret.len > 1)
ret.len--;
return ret;
}
int BigNum::operator %(const int & b) const    //大数对一个int类型的变量进行取模运算
{
int i,d=0;
for (i = len-1; i>=0; i--)
{
d = ((d * (MAXN+1))% b + a[i])% b;
}
return d;
}
BigNum BigNum::operator^(const int & n) const    //大数的n次方运算
{
BigNum t,ret(1);
int i;
if(n<0)
exit(-1);
if(n==0)
return 1;
if(n==1)
return *this;
int m=n;
while(m>1)
{
t=*this;
for( i=1;i<<1<=m;i<<=1)
{
t=t*t;
}
m-=i;
ret=ret*t;
if(m==1)
ret=ret*(*this);
}
return ret;
}
bool BigNum::operator>(const BigNum & T) const   //大数和另一个大数的大小比较
{
int ln;
if(len > T.len)
return true;
else if(len == T.len)
{
ln = len - 1;
while(a[ln] == T.a[ln] && ln >= 0)
ln--;
if(ln >= 0 && a[ln] > T.a[ln])
return true;
else
return false;
}
else
return false;
}
bool BigNum::operator >(const int & t) const    //大数和一个int类型的变量的大小比较
{
BigNum b(t);
return *this>b;
}

void BigNum::print()    //输出大数
{
int i;
cout << a[len - 1];
for(i = len - 2 ; i >= 0 ; i--)
{
cout.width(DLEN);
cout.fill('0');
cout << a[i];
}
cout << endl;
}
int main(void)
{
int i,n;
BigNum x[101];      //定义大数的对象数组
x[0]=1;
for(i=1;i<101;i++)
x[i]=x[i-1]*(4*i-2)/(i+1);
while(scanf("%d",&n)==1 && n!=-1)
{
x
.print();
}
}

附录：

#include <iostream>
#include <string>
using namespace std;

int compare(string str1, string str2)
{
//相等返回0，大于返回1，小于返回-1
if (str1.size()>str2.size()) return 1; //长度长的整数大于长度小的整数
else if (str1.size()<str2.size()) return -1;
else return str1.compare(str2); //若长度相等，则头到尾按位比较
}

string SUB_INT(string str1, string str2);
string ADD_INT(string str1, string str2) {//高精度加法
int sign = 1; //sign 为符号位
string str;
if (str1[0] == '-') {
if (str2[0] == '-') {
sign = -1;
str = ADD_INT(str1.erase(0, 1), str2.erase(0, 1));
}
else {
str = SUB_INT(str2, str1.erase(0, 1));
}
}
else {
if (str2[0] == '-') {
str = SUB_INT(str1, str2.erase(0, 1));
}
else { //把两个整数对齐，短整数前面加0补齐
string::size_type L1, L2;
int i;
L1 = str1.size();
L2 = str2.size();
if (L1<L2) {
for (i = 1; i <= L2 - L1; i++) str1 = "0" + str1;
}
else {
for (i = 1; i <= L1 - L2; i++) str2 = "0" + str2;
}
int int1 = 0, int2 = 0; //int2 记录进位
for (i = str1.size() - 1; i >= 0; i--) {
int1 = (int(str1[i]) - '0' + int(str2[i]) - '0' + int2) % 10;
int2 = (int(str1[i]) - '0' + int(str2[i]) - '0' + int2) / 10;
str = char(int1 + '0') + str;
}
if (int2 != 0) str = char(int2 + '0') + str;
}
}
//运算后处理符号位
if ((sign == -1) && (str[0] != '0')) str = "-" + str;
return str;
}

string SUB_INT(string str1, string str2) {//高精度减法
int sign = 1; //sign 为符号位
string str;
int i, j;
if (str2[0] == '-') {
str = ADD_INT(str1, str2.erase(0, 1));
}
else {
int res = compare(str1, str2);
if (res == 0) return "0";
if (res<0) {
sign = -1;
string temp = str1;
str1 = str2;
str2 = temp;
}
string::size_type tempint;
tempint = str1.size() - str2.size();
for (i = str2.size() - 1; i >= 0; i--) {
if (str1[i + tempint]<str2[i]) {
j = 1;
while (1) {//zhao4zhong1添加
if (str1[i + tempint - j] == '0') {
str1[i + tempint - j] = '9';
j++;
}
else {
str1[i + tempint - j] = char(int(str1[i + tempint - j]) - 1);
break;
}
}
str = char(str1[i + tempint] - str2[i] + ':') + str;
}
else {
str = char(str1[i + tempint] - str2[i] + '0') + str;
}
}
for (i = tempint - 1; i >= 0; i--) str = str1[i] + str;
}
//去除结果中多余的前导0
str.erase(0, str.find_first_not_of('0'));
if (str.empty()) str = "0";
if ((sign == -1) && (str[0] != '0')) str = "-" + str;
return str;
}

string MUL_INT(string str1, string str2) { //高精度乘法
int sign = 1; //sign 为符号位
string str;
if (str1[0] == '-') {
sign *= -1;
str1 = str1.erase(0, 1);
}
if (str2[0] == '-') {
sign *= -1;
str2 = str2.erase(0, 1);
}

int i, j;
string::size_type L1, L2;
L1 = str1.size();
L2 = str2.size();
for (i = L2 - 1; i >= 0; i--)
{
//模拟手工乘法竖式
string tempstr;
int int1 = 0, int2 = 0, int3 = int(str2[i]) - '0';
if (int3 != 0)
{
for (j = 1; j <= (int)(L2 - 1 - i); j++)
tempstr = "0" + tempstr;

for (j = L1 - 1; j >= 0; j--)
{
int1 = (int3*(int(str1[j]) - '0') + int2) % 10;
int2 = (int3*(int(str1[j]) - '0') + int2) / 10;
tempstr = char(int1 + '0') + tempstr;
}
if (int2 != 0) tempstr = char(int2 + '0') + tempstr;
}
str = ADD_INT(str, tempstr);
}

//去除结果中的前导0
str.erase(0, str.find_first_not_of('0'));
if (str.empty()) str = "0";
if ((sign == -1) && (str[0] != '0'))
str = "-" + str;
return str;
}

string DIVIDE_INT(string str1, string str2, int flag) { //高精度除法。flag==1时,返回商; flag==0时,返回余数
string quotient, residue; //定义商和余数
int sign1 = 1, sign2 = 1;
if (str2 == "0") { //判断除数是否为0
quotient = "ERROR!";
residue = "ERROR!";
if (flag == 1) return quotient;
else return residue;
}
if (str1 == "0") { //判断被除数是否为0
quotient = "0";
residue = "0";
}
if (str1[0] == '-') {
str1 = str1.erase(0, 1);
sign1 *= -1;
sign2 = -1;
}
if (str2[0] == '-') {
str2 = str2.erase(0, 1);
sign1 *= -1;
}
int res = compare(str1, str2);
if (res<0) {
quotient = "0";
residue = str1;
}
else if (res == 0) {
quotient = "1";
residue = "0";
}
else {
string::size_type L1, L2;
L1 = str1.size();
L2 = str2.size();
string tempstr;
tempstr.append(str1, 0, L2 - 1);
for (int i = L2 - 1; i<L1; i++) { //模拟手工除法竖式
tempstr = tempstr + str1[i];
tempstr.erase(0, tempstr.find_first_not_of('0'));//zhao4zhong1添加
if (tempstr.empty()) tempstr = "0";//zhao4zhong1添加
for (char ch = '9'; ch >= '0'; ch--) { //试商
string str;
str = str + ch;
if (compare(MUL_INT(str2, str), tempstr) <= 0) {
quotient = quotient + ch;
tempstr = SUB_INT(tempstr, MUL_INT(str2, str));
break;
}
}
}
residue = tempstr;
}
//去除结果中的前导0
quotient.erase(0, quotient.find_first_not_of('0'));
if (quotient.empty()) quotient = "0";
if ((sign1 == -1) && (quotient[0] != '0')) quotient = "-" + quotient;
if ((sign2 == -1) && (residue[0] != '0')) residue = "-" + residue;
if (flag == 1) return quotient;
else return residue;
}

string DIV_INT(string str1, string str2) {//高精度除法,返回商
return DIVIDE_INT(str1, str2, 1);
}

string MOD_INT(string str1, string str2) {//高精度除法,返回余数
return DIVIDE_INT(str1, str2, 0);
}

int main()
{
/*char ch;
string s1, s2, res;

while (cin >> s1 >> ch >> s2) {
switch (ch) {
case '+':res = ADD_INT(s1, s2); break;
case '-':res = SUB_INT(s1, s2); break;
case '*':res = MUL_INT(s1, s2); break;
case '/':res = DIV_INT(s1, s2); break;
case '%':res = MOD_INT(s1, s2); break;
default: break;
}
cout << res << endl;
}
return(0);*/

string s11, s22, res1;

s11 = "1";
s22 = "2";
for (int i = 1; i <= 1000; i++)
{
res1 = MUL_INT(s11, s22); //利用乘法做幂次运算！
s11 = res1;
}
cout << res1 << endl;
return(0);
}