大数运算模板

#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define MAXN 9999#define MAXSIZE 10#define DLEN 4class BigInt{private:    int a[500];    //可以控制大数的位数    int len;       //大数长度public:    BigInt(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数    BigInt(const int);       //将一个int类型的变量转化为大数    BigInt(const char*);     //将一个字符串类型的变量转化为大数    BigInt(const BigInt &);  //拷贝构造函数    BigInt &operator=(const BigInt &);   //重载赋值运算符，大数之间进行赋值运算    friend istream& operator>>(istream&,  BigInt&);   //重载输入运算符    friend ostream& operator<<(ostream&,  BigInt&);   //重载输出运算符    BigInt operator+(const BigInt &) const;   //重载加法运算符，两个大数之间的相加运算    BigInt operator-(const BigInt &) const;   //重载减法运算符，两个大数之间的相减运算    BigInt operator*(const BigInt &) const;   //重载乘法运算符，两个大数之间的相乘运算    BigInt operator/(const int   &) const;    //重载除法运算符，大数对一个整数进行相除运算    BigInt 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 BigInt & T)const;   //大数和另一个大数的大小比较    bool   operator<(const BigInt & T) const;    bool   operator==(const BigInt & T) const;    bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较    bool   operator<(const int &t) const;    bool   operator==(const int &t) const;    void print();       //输出大数};bool BigInt::operator==(const BigInt & T) const {    return !(*this > T) && !(T > *this);}bool BigInt::operator==(const int &t) const {    BigInt T = BigInt(t);    return *this == T;}bool BigInt::operator<(const BigInt & T) const {    return T > *this;}bool BigInt::operator<(const int &t) const {    return BigInt(t) > *this;}BigInt::BigInt(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;}BigInt::BigInt(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;    }}BigInt::BigInt(const BigInt & 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];}BigInt & BigInt::operator=(const BigInt & 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,  BigInt & 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,  BigInt& 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;}BigInt BigInt::operator+(const BigInt & T) const   //两个大数之间的相加运算{    BigInt 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;}BigInt BigInt::operator-(const BigInt & T) const   //两个大数之间的相减运算{    int i,j,big;    bool flag;    BigInt 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;}BigInt BigInt::operator*(const BigInt & T) const   //两个大数之间的相乘运算{    BigInt 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;}BigInt BigInt::operator/(const int & b) const   //大数对一个整数进行相除运算{    BigInt 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 BigInt::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;}BigInt BigInt::operator^(const int & n) const    //大数的n次方运算{    BigInt 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 BigInt::operator>(const BigInt & 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 BigInt::operator >(const int & t) const    //大数和一个int类型的变量的大小比较{    BigInt b(t);    return *this>b;}void BigInt::print()    //输出大数{    int i;    printf("%d", a[len-1]);    for (int i = len-2; i >= 0; --i) {        printf("%04d", a[i]);    }    puts("");}int main(){    int t;    char a[1000], b[1000];    BigInt ans;    scanf("%s%s",a,b);    BigInt A(a), B(b);    printf("%s + %s = ", a, b);    ans = A + B;    ans.print();    printf("%s - %s = ", a, b);    ans = A - B;    ans.print();    scanf("%d",&t);    printf("%s * %d = ", a, t);    ans = A * t;    ans.print();    printf("%s / %d = ", a, t);    ans = A / t;    ans.print();    printf("%s ^ %d = ", a, t);    ans = A ^ t;    ans.print();    printf("%s %% %d = ", a, t);    ans = A % t;    ans.print();    return 0;}

下面是用string类写的模板

两个大整数相加：

/*大整数加法调用方式:add(a, b);返回类型：string*/string add(string a, string b){    string s;    reverse(a.begin(), a.end());    reverse(b.begin(), b.end());    int i = 0;    int m, k = 0;    while(a[i] && b[i])    {        m = a[i] - '0' + b[i] - '0' + k;        k = m / 10;        s += (m % 10 + '0');        i++;    }    if(i == a.size())    {        while(i != b.size())        {            m = k + b[i] - '0';            k = m / 10;            s += m % 10 + '0';            i++;        }        if(k) s += k + '0';    }    else if(i == b.size())    {        while(i != a.size())        {            m = k + a[i] - '0';            k = m / 10;            s += m % 10 + '0';            i++;        }        if(k) s += k + '0';    }    reverse(s.begin(), s.end());    return s;}

两个大整数相减：

/*    大整数减法*/#include<cstdio>#include<cstring>#include<string>#include<iostream>using namespace std;string sub(string a, string b){    int i, j, k, s, flag = 1;    int tmpa[10000], tmpb[10000], c[10000];    string ans;    if(a.size() < b.size() || (a.size() == b.size() && a.compare(b) < 0))    {        string tmp = a;        a = b;        b = tmp;        flag = 0;    }    while(a.length() > b.length()) b = '0' + b;    int len = a.length();    for(i = 0; i < len; i++)    {        tmpa[i] = a[i] - '0';        tmpb[i] = b[i] - '0';    }    for(i = len - 1; i >= 0; i--)    {        if(tmpa[i] >= tmpb[i])            c[i] = tmpa[i] - tmpb[i];        else        {            c[i] = 10 + tmpa[i] - tmpb[i];            tmpa[i-1]--;        }    }    for(i = 0; i < len - 1; i++)        if(c[i] != 0)            break;    for(j = i; j < len; j++)        ans = ans + (char)(c[j] + '0');    if(!flag)        ans = '-' + ans;    return ans;}int main(){    string a, b;    while(cin >> a >> b)    {        cout << sub(a, b) << endl;    }    return 0;}

大整数乘以一个小整数：

/*    大整数乘以小整数（可以用整型变量表示的整数）*/string multi(string a, int k){    if(k == 0) return "0";    int len = a.length(), carry = 0;    reverse(a.begin(), a.end());    for(int i = 0; i < len; i++)    {        int s = (a[i] - '0') * k + carry;        a[i] = s % 10 + '0';        carry = s / 10;    }    while(carry != 0)    {        a = a + (char)(carry % 10 + '0');        carry /= 10;    }    reverse(a.begin(), a.end());    return a;}

两个不含前导0的大整数相乘：

/*    两个不含前导0的大整数相乘*/#include<string>#include<iostream>#include<algorithm>using namespace std;string multi_string_int(string a, int k){    if(k == 0) return "0";    int len = a.length(), carry = 0;    reverse(a.begin(), a.end());    for(int i = 0; i < len; i++)    {        int s = (a[i] - '0') * k + carry;        a[i] = s % 10 + '0';        carry = s / 10;    }    while(carry != 0)    {        a = a + (char)(carry % 10 + '0');        carry /= 10;    }    reverse(a.begin(), a.end());    return a;}string add(string a, string b){    string s;    reverse(a.begin(), a.end());    reverse(b.begin(), b.end());    int i = 0;    int m, k = 0;    while(a[i] && b[i])    {        m = a[i] - '0' + b[i] - '0' + k;        k = m / 10;        s += (m % 10 + '0');        i++;    }    if(i == a.size())    {        while(i != b.size())        {            m = k + b[i] - '0';            k = m / 10;            s += m % 10 + '0';            i++;        }        if(k) s += k + '0';    }    else if(i == b.size())    {        while(i != a.size())        {            m = k + a[i] - '0';            k = m / 10;            s += m % 10 + '0';            i++;        }        if(k) s += k + '0';    }    reverse(s.begin(), s.end());    return s;}string multi_string_string(string a, string b){    string ans = "";    for(int i = a.size() - 1; i >= 0; i--)    {        string tmp = multi_string_int(b, a[i] - '0');        for(int j = 0; j < a.size() - 1 - i; j++)            tmp += '0';        ans = add(ans, tmp);    }    return ans;}int main(){    string a, b;    while(cin >> a >> b)    {        cout << multi_string_string(a, b) << endl;    }    return 0;}

求大整数除以一个小整数的商：

string division(string str, int x) {    string ans = "";    int len = str.length();    int y = 0;    for(int i = 0; i < len; i++) {        ans += char((y * 10 + (str[i] - '0')) / x + '0');        y = (y * 10 + (str[i] - '0')) % x;    }    while(*(ans.begin()) == '0' && ans.size() > 1) ans.erase(ans.begin());    return ans;}

求一个大整数对一个小整数的余数：

int Mod(string str, int x) {    int len = str.length();    int y = 0;    for(int i = 0; i < len; i++)        y = (y * 10 + (str[i] - '0')) % x;    return y;}

两个大数除法：

/* 大数除法 -- 高精度除高精度  */#include <iostream>#include <string>#include <cstring>using namespace std;#define N 2000/*1.  a.size < b.size 返回-12.  a.size == b.size && a - b < 0 返回-13.  a.size == b.size && a - b == 0 返回0*/// 判断a.size 与 b.size的关系，以及做减法int Judge(char a[], int a1, char b[], int b1) {    int i;    if(a1 < b1) return -1; // a.size < b.size    bool flag = false;    if(a1 == b1) { //a.size == b.size && a < b        for(i = a1 - 1; i >= 0; i--) {            if(a[i] > b[i])                flag = true;            else if(a[i] < b[i]) {                if(!flag) return -1;            }        }    }    for(i = 0; i < a1; i++) { //前提b中b1--a1部分必须为'0'        a[i] = a[i] - b[i] + 48; //'0'的ASCII为48        if((a[i] - '0') < 0) {            a[i] = a[i] + 10;            a[i+1] = a[i+1] - 1;        }    }    for(i = a1 - 1; i >= 0; i--) { // 返回被除数的长度        if(a[i] != '0')            return (i + 1);    }    return 0; //a.size==b.size&&a==b的情况}string division(string a, string b) {    char x1[N], x2[N];    int ans[N];    int a_len, b_len, i, j;    a_len = a.length();    b_len = b.length();    // 初始化    memset(x1, '0', sizeof(x1));    memset(x2, '0', sizeof(x2));    memset(ans, 0, sizeof(ans));    for(i = a_len - 1, j = 0; i >= 0; i--)        x1[j++] = a[i];    for(i = b_len - 1, j = 0; i >= 0; i--)        x2[j++] = b[i];    // 分析部分    if(a_len < b_len) return "0";    int temp_len = Judge(x1, a_len, x2, b_len);    if(temp_len < 0) return "0";    if(temp_len == 0) return "1";    ans[0]++; // 减掉1位， 商加1    int ntimes = temp_len - b_len;    if(ntimes < 0) return "1";    // 扩充数位， 加快减法    else if(ntimes > 0) {        for(i = temp_len - 1; i >= 0; i--) {            if(i >= ntimes)                x2[i] = x2[i - ntimes];            else                x2[i] = '0';        }    }    b_len = temp_len;    // 加快除法的部分    for(j = 0; j <= ntimes; j++) {        int ntemp;        while((ntemp = Judge(x1, temp_len, x2 + j, b_len - j)) >= 0) {            temp_len = ntemp;            ans[ntimes - j]++;        }    }    for(i = 0; i < N; i++) {        if(ans[i] >= 10) {            ans[i+1] += ans[i] / 10;            ans[i] %= 10;        }    }    // 返回string类型    int k = N - 1;    string c = "";    while(ans[k] == 0 && k > 0) k--;    for(i = k; i >= 0; i--)        c+= (ans[i] + '0');    return c;}int main(){    string a, b;    while(cin >> a >> b) {        cout << division(a, b) << endl;    }    return 0;}