大数运算

来源:互联网 发布:java新建package 编辑:程序博客网 时间:2024/05/01 07:42
 
#include <iostream>#include <string>using namespace std;inline int compare(string str1, string str2) {    if(str1.size() > str2.size()) //长度长的整数大于长度小的整数        return 1;    else if(str1.size() < str2.size())        return -1;    else        return str1.compare(str2); //若长度相等,从头到尾按位比较,compare函数:相等返回0,大于返回1,小于返回-1}string ADD_INT(string str1, string str2) {//高精度加法    string SUB_INT(string str1, string str2);    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    string MUL_INT(string str1, string str2);    int sign = 1; //sign 为符号位    string str;    int i;    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]) {                str1[i + tempint - 1] = char(int(str1[i + tempint - 1]) - 1);                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 --) {  //实现手工乘法        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);        //模拟手工除法        for(int i = l2 - 1; i < l1; i++) {            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 >> ch) {        cin >> s1 >> 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);}