大整数类加减乘除的简单实现——C++

编程题＃4：大整数的加减乘除

来源: POJ 

总时间限制: 

1000ms 

内存限制: 

65536kB

描述

给出两个正整数以及四则运算操作符（+ - * /），求运算结果。

输入

第一行：正整数a，长度不超过100
第二行：四则运算符o，o是“+”，“-”，“*”，“/”中的某一个
第三行：正整数b，长度不超过100

保证输入不含多余的空格或其它字符

输出

一行：表达式“a o b”的值。

补充说明：
1. 减法结果有可能为负数
2. 除法结果向下取整
3. 输出符合日常书写习惯，不能有多余的0、空格或其它字符

样例输入

9876543210+9876543210

样例输出

19753086420

这次的大整数加减乘除的基本思想是用字符串构造一个类，记录十进制的数字。用竖式计算实现加和减。接下来要注意，对于乘法，因为只有+/- 可以调用，若要求求得a*b，仅仅用a次循环的+=b是极慢的。所以，要用递归来实现，若用res记录结果，第一次res=b，

第n次可使res=res+res;这样res的值以二次方增长，大大节省时间。除法也是如此。

</pre><pre name="code" class="cpp"><pre name="code" class="cpp">#include <iostream>#include <string.h>#include <string>#include <stdlib.h>#include <sstream>#include <stdio.h>using namespace std;void itoa1(int a, char* b, int c);long long atoi1(const char s[])                            //atoi{long long temp = 0;const char *ptr = s;if (*s == '-' || *s == '+'){s++;}while (*s != 0){if ((*s < '0') || (*s > '9')){break;}temp = temp * 10LL + (long long)(*s - '0');s++;}if (*ptr == '-') {temp = -temp;}return temp;}class longint {string a;public:longint(string);longint();longint(const char*);string operator=(const char*);string operator+(longint);string operator-(longint);string operator*(longint);string operator/(longint);};void itoa1(int n, char *str, int c)                         //itoa {char buf[10] = "";int i = 0;int len = 0;int temp = n < 0 ? -n : n;if (n == 0) {str[0] = '0';str[1] = 0;return;}if (str == NULL){return;}while (temp){buf[i++] = (temp % 10) + '0';temp = temp / 10;}len = n < 0 ? ++i : i;  //如果n是负数，则多需要一位来存储负号str[i] = 0;            //末尾是结束符0while (1){i--;if (buf[len - i - 1] == 0){break;}str[i] = buf[len - i - 1];  //把buf数组里的字符拷到字符串}if (i == 0){str[i] = '-';          //如果是负数，添加一个负号}}int cmpl(string f, string l);                      int main() {string k;char o = 0; string kk;cin >> k >> o >> kk;longint a(k), b(kk);switch (o){case '+':cout << a + b << endl; break;case '-':cout << a - b << endl; break;case '*':cout << a * b << endl; break;case '/':cout << a / b << endl; break;default:break;}/*char tes[30] = {}; long long k = 0;while (cin >> k) {itoa1(k, tes, 10);cout<<tes<< endl << endl;}*/cin.get();cin.get();return 0;}int cmpl(string f, string l) {                                             //用于比较字符串大小if (f.length() > l.length()) return 1;else if (f.length() < l.length()) return -1;else {for (int i = 0; i < f.length(); i++) {if (f[i] > l[i])return 1;if (l[i] > f[i])return -1;}}return 0;}longint::longint(string b) {a = b;}longint::longint(){}longint::longint(const char *b){a = b;}string longint::operator=(const char * b){a = b;return a;}string longint::operator+(longint b){int maxlen = 0;string maxN;string minN;if (a.length() >= b.a.length()) {maxlen = a.length();maxN = "0";maxN += a;minN.resize(a.length() - b.a.length() + 1, '0');minN += b.a;}else {maxlen = b.a.length();maxN = "0";maxN += b.a;minN.resize(b.a.length() - a.length() + 1, '0');minN += a;}bool ad = 0;string Sr; Sr.resize(maxlen + 1, '0');for (int i = maxlen; i >= 0; i--) {int res = 0;if (ad) {res += 1;ad = 0;}char in1[2] = { maxN[i],'\0' };char in2[2] = { minN[i],'\0' };res += atoi1(in1) + atoi1(in2);if (res > 9) {ad = 1;char temp[3] = {}; itoa1(res, temp, 10);Sr[i] = temp[1];}else {char temp[2] = {}; itoa1(res, temp, 10);Sr[i] = temp[0];}}if (Sr[0] == '0') {Sr.erase(0, 1);}return Sr;}string longint::operator-(longint b){int maxlen = 0; bool nag = 0;string maxN;string minN;if (cmpl(a, b.a) == 1) {maxlen = a.length();maxN = "0";maxN += a;minN.resize(a.length() - b.a.length() + 1, '0');minN += b.a;}else if (cmpl(a, b.a) == -1) {nag = 1;maxlen = b.a.length();maxN = "0";maxN += b.a;minN.resize(b.a.length() - a.length() + 1, '0');minN += a;}else return "0";bool ad = 0;string Sr; Sr.resize(1, '0'); Sr.resize(maxlen + 1, '0');int res = 0;for (int i = maxlen; i >= 0; i--) {res = 0;if (ad) {res -= 1;ad = 0;}char in1[2] = { maxN[i],'\0' };char in2[2] = { minN[i],'\0' };if (in1[0] + res< in2[0]) {res += 10 + atoi1(in1) - atoi1(in2);ad = 1;}else res += atoi1(in1) - atoi1(in2);char temp[2] = {}; itoa1(res, temp, 10);Sr[i] = temp[0];}while (Sr[0] == '0'&&Sr[1] == '0') {Sr.erase(0, 1);}if (nag) {Sr[0] = '-';}else {Sr.erase(0, 1);}if (Sr[0] == 0) {Sr.erase(0, 1);}return Sr;}string longint::operator*(longint b){longint maxN;longint minN;if (a == "0" || b.a == "0")return "0";else if (cmpl(a, b.a) == 1) {maxN = a;minN = b.a;}else {maxN = b.a;minN = a;}longint times = minN;longint Sr;longint i = "1";Sr = maxN;/*for (; cmpl(i+i, times.a) != 1; i = i + i) {   // i<timesSr = Sr + Sr;}*/for (; cmpl(i.a, times.a) ==-1;) {                              //类似递归的循环longint k = "1";longint Sr1 = maxN;for (; cmpl(k+k, times - i) ==-1; k = k + k) {Sr1 = Sr1 + Sr1;}i = k + i;Sr = Sr + Sr1;}return Sr.a;}string  longint::operator/(longint b){longint maxN = a;longint minN = b.a;longint i = "0";longint k = "1";longint Sr1 = minN;for (; cmpl(((minN + minN)), (maxN.a.c_str())) != 1;) {k = "1";for (; cmpl((Sr1 + Sr1), maxN.a) != 1; k = k*"2") {Sr1 = Sr1 + Sr1;}i = k + i;maxN = maxN - Sr1;Sr1 = minN;k = "2";}for (; maxN.a[0] != '-'; i = i + "1") {                           <span style="font-family: Arial, Helvetica, sans-serif;">//类似递归的循环</span>maxN = maxN - minN;}i = i - "1";return i.a;}/*741269586512                               // 一些测试数据*98456247865456765156165465786156467/5465781545478248789151687984156*564871123467868465156478654165468421-1564687811548785431223456748951487894146576543215646+512312316578451231321568687894123467984512789/56841.7298262214470804063148482.27559582927035448314667455226163.10385551574.-156384129590092001467661465.540746110599310888564313326.15463427774100019*/

另外，大整数也可以用链表、数组等来实现。

最后，提醒来找代码交作业的同学，不要直接拷贝啊！！！！这样没有任何好处！