c++的正整数高精度加减乘除

  1 //
  2 //  main.cpp
  3 //  正整数高精度运算
  4 //
  5 //  Created by ashley on 14-11-9.
  6 //  Copyright (c) 2014年 ashley. All rights reserved.
  7 //
  8 
  9 #include <iostream>
 10 #include <string>
 11 using namespace std;
 12 
 13 string clearZeros(string data)
 14 {
 15     if (data[0] == ‘0‘) {
 16         int key = (int) data.length() - 1;
 17         for (int i = 0; i < data.length(); i++) {
 18             if (data[i] != ‘0‘) {
 19                 key = i;
 20                 break;
 21             }
 22         }
 23         data.erase(0, key);
 24     }
 25     if (data == "") {
 26         data = "0";
 27     }
 28     return data;
 29 }
 30 
 31 //对位操作
 32 void countPoint(string &operand1, string &operand2)
 33 {
 34     while (operand1.length() < operand2.length()) {
 35         operand1 = "0" + operand1;
 36     }
 37     while (operand1.length() > operand2.length()) {
 38         operand2 = "0" + operand2;
 39     }
 40 }
 41 
 42 //判断大小
 43 bool bigger(string operand1, string operand2)
 44 {
 45     return operand1 >= operand2;
 46 }
 47 
 48 string addition(string addent, string adder)
 49 {
 50     //先对位，在加数和被加数前面适当补0，使他们包含相同的位数
 51     countPoint(addent, adder);
 52     //前面再补一个0，确定和的最多位数
 53     addent = "0" + addent;
 54     adder = "0" + adder;
 55     //从低位开始，对应位相加，结果写进被加数中，如果有进位，直接给被加数前一位加1
 56     for (int i = (int) addent.length() - 1; i > 0; i--) {
 57         addent[i] = addent[i] + adder[i] - 48;
 58         if (addent[i] > ‘9‘) {
 59             addent[i] = addent[i] - 10;
 60             addent[i - 1] = addent[i - 1] + 1;
 61         }
 62     }
 63     return clearZeros(addent);
 64 }
 65 
 66 string subtraction(string subtrahend, string subtractor)
 67 {
 68     //先对位，在减数和被减数前面适当补0，使他们包含相同的位数
 69     countPoint(subtrahend, subtractor);
 70     //判断被减数和减数谁大，保证被减数大于减数
 71     if (bigger(subtrahend, subtractor)) {
 72         subtrahend[0] = subtrahend[0] - subtractor[0] + 48;
 73         for (int i = 1; i < (int)subtrahend.length(); i++) {
 74             if (subtrahend[i] >= subtractor[i]) {
 75                 subtrahend[i] = subtrahend[i] - subtractor[i] + 48;
 76             } else {
 77                 subtrahend[i] = subtrahend[i] - subtractor[i] + 10 + 48;
 78                 subtrahend[i - 1]--;
 79             }
 80         }
 81     } else {
 82         subtrahend = ‘-‘ + subtraction(subtractor, subtrahend);
 83     }
 84     return subtrahend;
 85 }
 86 
 87 string multiplication(string multiplicand, string multiplier)
 88 {
 89     string result = "0";
 90     for (int i = (int)multiplier.length() - 1; i >= 0 ; i--) {
 91         for (char c = ‘1‘; c <= multiplier[i]; c++) {
 92             result = addition(result, multiplicand);
 93         }
 94         multiplicand = multiplicand + "0";
 95     }
 96     return clearZeros(result);
 97 }
 98 
 99 // 试商法
100 string division(string dividend, string divisor)
101 {
102     // 存放商
103     string result;
104     // 存放余数
105     string remains;
106     for (int i = 0; i < (int)dividend.length(); i++) {
107         remains = remains + dividend[i];
108         result = result + "0";
109         // 从1往上试
110         while (bigger(remains, result)) {
111             cout << result << "-----------" << remains << endl;
112             result[result.length() - 1]++;
113             remains = subtraction(remains, divisor);
114         }
115     }
116     return clearZeros(result);
117 }
118 int main(int argc, const char * argv[])
119 {
120     string a, b;
121     int tests;
122     cin >> tests;
123     while (tests--) {
124         cin >> a >> b;
125         //正整数高精度加法，从低位开始
126         //cout << addition(a, b) << endl;
127         //正整数高精度减法，从高位开始
128         //cout << subtraction(a, b) << endl;
129         //正整数高精度乘法，将乘法转换为加法进行运算
130         //cout << multiplication(a, b) << endl;
131         cout << division(a, b) << endl;
132         //正整数高精度除法
133 
134     }
135     return 0;
136 }