大整數類加減乘除的簡單實現——C++
阿新 • • 發佈：2019-01-27
程式設計題＃4：大整數的加減乘除
- 總時間限制:
- 1000ms
- 記憶體限制:
- 65536kB
- 描述
-
給出兩個正整數以及四則運算操作符（+ - * /），求運算結果。
- 輸入
- 第一行：正整數a，長度不超過100
第二行：四則運算子o，o是“+”，“-”，“*”，“/”中的某一個
第三行：正整數b，長度不超過100
保證輸入不含多餘的空格或其它字元 - 輸出
- 一行：表示式“a o b”的值。
補充說明：
1. 減法結果有可能為負數
2. 除法結果向下取整
3. 輸出符合日常書寫習慣，不能有多餘的0、空格或其它字元 - 樣例輸入
-
9876543210 + 9876543210
- 樣例輸出
-
19753086420
這次的大整數加減乘除的基本思想是用字串構造一個類，記錄十進位制的數字。
第n次可使res=res+res;這樣res的值以二次方增長，大大節省時間。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是負數，則多需要一位來儲存負號 str[i] = 0; //末尾是結束符0 while (1) { i--; if (buf[len - i - 1] == 0) { break; } str[i] = buf[len - i - 1]; //把buf數組裡的字元拷到字串 } if (i == 0) { str[i] = '-'; //如果是負數，新增一個負號 } } 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) { //用於比較字串大小 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<times Sr = Sr + Sr; }*/ for (; cmpl(i.a, times.a) ==-1;) { //類似遞迴的迴圈 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;">//類似遞迴的迴圈</span> maxN = maxN - minN; }i = i - "1"; return i.a; } /* 741269586512 // 一些測試資料 * 984562478654 56765156165465786156467 / 54657815454782 48789151687984156 * 56487112346786 8465156478654165468421 - 15646878115487854312234567 48951487894146576543215646 + 5123123165784512313215686 87894123467984512789 / 5684 1.729826221447080406314848 2.2755958292703544831466745522616 3.1038555157 4.-15638412959009200146766146 5.54074611059931088856431332 6.15463427774100019 */
另外，大整數也可以用連結串列、陣列等來實現。