1034 有理数四则运算 (20 分)(c语言)(测试点三)

本题要求编写程序，计算 2 个有理数的和、差、积、商。

输入格式：

输入在一行中按照 a1/b1 a2/b2 的格式给出两个分数形式的有理数，其中分子和分母全是整型范围内的整数，负号只可能出现在分子前，分母不为 0。

输出格式：

分别在 4 行中按照 有理数1 运算符有理数2 = 结果 的格式顺序输出 2 个有理数的和、差、积、商。注意输出的每个有理数必须是该有理数的最简形式 k a/b，其中 k 是整数部分，a/b 是最简分数部分；若为负数，则须加括号；若除法分母为 0，则输出 Inf。题目保证正确的输出中没有超过整型范围的整数。

输入样例 1：

2/3 -4/2

结尾无空行

输出样例 1：

2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)

结尾无空行

输入样例 2：

5/3 0/6

结尾无空行

输出样例 2：

1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf

结尾无空行

#include<stdio.h>
#include<math.h>
#pragma warning (disable:4996)
long long  gcd(long long  a, long long  b);
void fun(long long  a, long long  b);
int main() {long long a1, b1, a2, b2;scanf("%lld/%lld %lld/%lld", &a1, &b1, &a2, &b2);fun(a1, b1); printf(" + "); fun(a2, b2); printf(" = "); fun(a1 * b2 + a2 * b1, b1 * b2); putchar('\n');fun(a1, b1); printf(" - "); fun(a2, b2); printf(" = "); fun(a1 * b2 - a2 * b1, b1 * b2); putchar('\n');fun(a1, b1); printf(" * "); fun(a2, b2); printf(" = "); fun(a1*a2, b1 * b2); putchar('\n');fun(a1, b1); printf(" / "); fun(a2, b2); printf(" = "); fun(a1 * b2, a2 * b1);return 0;}
long long  gcd(long long a, long long b) {return b == 0 ? a : gcd(b, a % b);
}
void fun(long long a, long long b) {if (a * b == 0) {printf("%s", a == 0 ? "0" : "Inf");return;}int   flag = a > 0 && b < 0 || a < 0 && b>0;printf("%s", flag ? "(-" : "");a = abs(a); b = abs(b);int k = a / b;if (k != 0) {printf("%d", k);}if (a % b == 0) {printf("%s", flag ? ")" : "");return;}printf("%s", k != 0 ? " " : "");long long  yueshu = gcd(a, b);a /= yueshu; b /= yueshu;printf("%lld/%lld%s",a%b,b, flag ? ")" : "");return;
}

这题目很神奇,我用c编译过不了题目中的测试点3,4,但是我改用成c++编译器,结果就通过了

这题的思路,其实就是分解,不过我最开始分解错了,我把它分解成了加减乘除四个,然后特别恶心,其实这道题因为就是两个数以内的加减乘除,所以其实可以直接把它们做成约分就行,然后一个个进行处理,还有注意下首末的括号,还有个重要的思路就是把负数换成正数,不然趋0取余可以把你逼疯

#include<iostream>
#include<string>
using namespace std;
#include<cmath>
#include<algorithm>
#pragma warning (disable:4996)
long long gcd(long long  a, long long b);
void fun(long long a, long long b);
int main() {long long  a1, b1, a2, b2;scanf("%lld/%lld %lld/%lld", &a1, &b1, &a2, &b2);fun(a1, b1); printf(" + "); fun(a2, b2); printf(" = "); fun(a1 * b2 + a2 * b1, b1 * b2); cout << "\n";fun(a1, b1); printf(" - "); fun(a2, b2); printf(" = "); fun(a1 * b2 - a2 * b1, b1 * b2); cout << "\n";fun(a1, b1); printf(" * "); fun(a2, b2); printf(" = "); fun(a1 * a2, b1 * b2); cout << "\n";fun(a1, b1); printf(" / "); fun(a2, b2); printf(" = "); fun(a1 * b2, a2 * b1); cout << "\n";
}
long long gcd(long long  a, long long b) {if (b == 0)return a;else gcd(b, a % b);
}
void fun(long long  a, long long  b) {if (a * b == 0) {cout << (a == 0 ? "0" : "Inf");return;}bool flag = a > 0 && b < 0 || a < 0 && b>0;if (flag) {cout << "(-";a = abs(a);b = abs(b);}long long  zhengshu = a / b;if (zhengshu != 0) {cout << zhengshu;}if (a % b == 0) {cout << (flag == 1 ? ")" : "");return;}else{cout << (zhengshu != 0 ? " " : "");long long  yueshu = gcd(a, b);a /= yueshu; b /= yueshu;a %= b;printf("%lld/%lld", a, b);cout << (flag == 1 ? ")" : "");}return;}

第二次写这道题.发现如果显示答案错误可以从输入入手,我第二次写这道题发现如果scanf;是%d的输入的话,答案会显示错误,应该是被截断的原因

第三次写,秒杀

#include<iostream>
using namespace std;
#include<string>
#include<algorithm>
#pragma warning (disable:4996)
#include <climits>
#include <vector>
#include<stack>
long long  f(long long  a, long long b);
void fun1(long long a, long long b);
int main() {long long a1, b1, a2, b2;scanf("%lld/%lld %lld/%lld", &a1, &b1, &a2, &b2);fun1(a1, b1); printf(" + "); fun1(a2, b2); printf(" = "); fun1(a1 * b2 + a2 * b1, b1 * b2); putchar('\n');fun1(a1, b1); printf(" - "); fun1(a2, b2); printf(" = "); fun1(a1 * b2 - a2 * b1, b1 * b2); putchar('\n');fun1(a1, b1); printf(" * "); fun1(a2, b2); printf(" = "); fun1(a1 * a2, b1 * b2); putchar('\n');fun1(a1, b1); printf(" / "); fun1(a2, b2); printf(" = "); fun1(a1 * b2, a2 * b1); putchar('\n');
}
void fun1(long long  a, long long b) {if (a * b == 0) {printf("%s", a == 0 ? "0" : "Inf");return;}bool flag = a > 0 && b < 0 || a < 0 && b>0;if (flag) {cout << "(-";a = abs(a);b = abs(b);}long long zhengshu = a / b;if (zhengshu != 0)cout << zhengshu;if (zhengshu != 0 && a % b != 0)cout << " ";if (a % b != 0) {long long gongyuehsu = f(a, b);a /= gongyuehsu;b /= gongyuehsu;cout << a % b << "/" << b;}if (flag)cout << ")";return;}
long long  f(long long  a,long long  b) {return b == 0 ? a : f(b, a % b);
}

哦,忽然记起来我写这道题的时候有个编译错误卡了我挺久的下面贴一下

严重性代码说明项目文件行禁止显示状态
错误 LNK2019 无法解析的外部符号 "__int64 __cdecl f(int,int)" (?f@@YA_JHH@Z)，函数 "void __cdecl fun1(__int64,__int64)" (?fun1@@YAX_J0@Z) 中引用了该符号 Project33 C:\Users\18960\source\repos\Project33\源.obj 1

我用vs2019的时候出了这个错误,然后查了下发现是我函数声明的时候把f函数形参声明成了int 类型,当时我在另一个函数里的时候调用时使用的是long long类型,然后发现这种错误是你在调用函数的时候,你声明的函数和实际调用的函数形参列表不一样导致的

long long  f(int a, int  b);
long long  f(long long a,long long  b) {return b == 0 ? a : f(b, a % b);
}

就上面这种