高精度计算1

高精度计算在最近几天上去，现贴上的只有高精度阶乘、加法、减法，加减法支持负数及小数，阶乘最多支持10000位，要增加请更改a数组的大小，加减法最多支持100位，若要修改，修改MAXN的大小

// 趣味C.cpp : 定义控制台应用程序的入口点。// 高精度的总结~#include <stdio.h>#include <string.h>#include <stdlib.h>#include <memory.h>#define MAXN 100void addition(char *num1, char *num2);void Subtract(char *num1, char *num2);// 阶乘void fact(int n){int a[10000];int nBit = 1;// 位数int temp;// 阶数的任一元素与临时结果的某位的乘积结果int i, j, carry;// 进位// 从1*2开始a[0] = 1;for (i = 2; i <= n; i++){carry = 0;// 每次算完都把进位清零for (j = 0; j < nBit; j++)// 按被乘数的位数来循环{temp = a[j] * i + carry;a[j] = temp % 10;carry = temp / 10;// 存进位}// 如果有进位digit位数就相加while (carry){++nBit;a[nBit - 1] = carry % 10;// 存进位carry /= 10;// 看还能不能进位}}printf("%d! = ", n);for (j = nBit; j >= 1; j--)printf("%d", a[j-1]);printf("/n/n");}// 初始化数据void InitData(char *num1, char *num2){// 计算小数点的位置size_t PosDotNum1 = 0, PosDotNum2 = 0;int i, diff;for (i = 0; num1[i] != '.' && num1[i] != '/0'; i++)PosDotNum1++;if (num1[i] == '/0'){num1[i] = '.';num1[i + 1] = '/0';}for (i = 0; num2[i] != '.' && num2[i] != '/0'; i++)PosDotNum2++;if (num2[i] == '/0'){num2[i] = '.';num2[i + 1] = '/0';}// 使两个数的小数点对齐if (PosDotNum1 < PosDotNum2){diff = PosDotNum2 - PosDotNum1;for (i = strlen(num1); i >= 0; i--)num1[i + diff] = num1[i];// 前导补0for (i = 0; i < diff; i++)num1[i] = '0';} else if (PosDotNum2 < PosDotNum1){diff = PosDotNum1 - PosDotNum2;for (i = strlen(num2); i >= 0; i--)num2[i + diff] = num2[i];// 前导补0for (i = 0; i < diff; i++)num2[i] = '0';}// 后继补0for (i = strlen(num1); i < MAXN - 1; i++)num1[i] = '0';num1[MAXN - 1] = '/0';for (i = strlen(num2); i < MAXN - 1; i++)num2[i] = '0';num2[MAXN - 1] = '/0';}// 打印结果void PrintResult(char *num){int i, n;for (i = 0; num[i] == '0'; i++);// 去除后继零for (n = MAXN - 1; num[n] == '0'; n--);if (-1 == n){printf("0/n");return;}if ('.' == num[n])num[n] = '/0';elsenum[n + 1] = '/0';if ('.' == num[i])i--;for (; num[i] != '/0'; i++)printf("%c", num[i]);printf("/n");}// 高精度加法void addition(char *num1, char *num2){char result[MAXN + 1];int i, temp;if ('-' == num1[0] && num2[0] != '-'){Subtract(num2, num1 + 1);return;}else if (num1[0] != '-' && num2[0] == '-'){Subtract(num1, num2 + 1);return;}else if (num1[0] == '-' && num2[0] == '-'){printf("-");addition(num1 + 1, num2 + 1);return;}// 初始化两个加数与和InitData(num1, num2);memset(result, '0', sizeof(result) / sizeof(char));result[MAXN] = '/0';// 计算和for (i = MAXN - 2; i >= 0; i--){if ('.' == num1[i]){result[i + 1] = '.';continue;}temp = (result[i + 1] - '0') + (num1[i] - '0') + (num2[i] - '0');result[i + 1] = temp % 10 + '0';if ('.' == num1[i - 1] && (temp / 10))result[i - 1] += temp / 10;result[i] += temp / 10;}// 打印结果PrintResult(result);}// 高精度减法void Subtract(char *num1, char *num2){char result[MAXN + 1];char temp1[MAXN], temp2[MAXN];int i, temp;if (num1[0] == '-' && num2[0] != '-'){printf("-");addition(num1 + 1, num2);return;}else if (num1[0] != '-' && num2[0] == '-'){strcpy(temp2, num2 + 1);addition(num1, temp2);return;}else if (num1[0] == '-' && num2[0] == '-'){strcpy(temp1, num1 + 1);strcpy(temp2, num2 + 1);Subtract(temp1, temp2);return;}// 初始化两个减数与差InitData(num1, num2);memset(result, '0', sizeof(result) / sizeof(char));result[MAXN] = '/0';if (strcmp(num1, num2) < 0){printf("-");Subtract(num2, num1);return;}// 计算差for (i = MAXN - 2; i >= 0; i--){if ('.' == num1[i]){result[i + 1] = '.';continue;}temp = (num1[i] - '0') - (num2[i] - '0');if (temp < 0 && i > 0){temp += 10;if ('.' == num1[i - 1])num1[i - 2]--;elsenum1[i - 1]--;}result[i + 1] = temp + '0';}// 打印计算结果PrintResult(result);}// 高精度乘法计算void multiply(char *num1, char *num2){}int main(int argc, char* argv[]){int sel;int n;char num1[MAXN], num2[MAXN];printf("===========================================/n");printf(" 欢迎使用高精度计算器 Dos版 VS.JQCTBU/n");printf("===========================================/n");while (1){printf("/n/n1.阶乘/n");printf("2.加法/n");printf("3.减法/n");printf("4.乘法/n");printf("5.除法/n");printf("6.多项式计算（简单）/n");printf("7.多项式计算（高精度）/n");printf("0.退出/n/n");printf("请输入你的选择：");scanf("%d", &sel); num1[0] = num2[0] = '/0';switch (sel){case 1:printf("/n请输入一个数：/n");scanf("%d", &n);fact(n);break;case 2:printf("/n请输入两个数/n第一个数：");scanf("%s", num1);printf("第二个数：");scanf("%s", num2);printf("%s + %s = /n", num1, num2);addition(num1, num2);break;case 3:printf("/n请输入两个数/n被减数：");scanf("%s", num1);printf("减数：");scanf("%s", num2);printf("%s - %s = /n", num1, num2);Subtract(num1, num2);break;case 4:printf("/n请输入两个需要相乘的数/n第一个数：");scanf("%s", num1);printf("第二个数：");scanf("%s", num2);multiply(num1, num2);break;case 0:exit(0);break;}}return 0;}