标签：amp 除法 编写 pre long using code mes show
本题要求编写程序，计算 2 个有理数的和、差、积、商。
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。
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。
2/3 -4/2
2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)
5/3 0/6
1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf#include<cstdio> #include<algorithm> using namespace std; typedef long long ll; struct Fraction{ ll up,down; }a,b; ll gcd(ll a, ll b){ return b == 0?a:gcd(b,a%b); } Fraction reduction(Fraction result){ if(result.down < 0){ result.down = -result.down; result.up = -result.up; } if(result.up == 0) result.down = 1; int d = gcd(abs(result.up),abs(result.down)); result.up /= d; result.down /= d; return result; } Fraction add(Fraction f1,Fraction f2){ Fraction result; result.up = f1.up*f2.down + f1.down * f2.up; result.down = f1.down*f2.down; return result; } Fraction minu(Fraction f1,Fraction f2){ Fraction result; result.up = f1.up*f2.down - f1.down * f2.up; result.down = f1.down*f2.down; return result; } Fraction multi(Fraction f1,Fraction f2){ Fraction result; result.up = f1.up*f2.up; result.down = f1.down*f2.down; return result; } Fraction divide(Fraction f1,Fraction f2){ Fraction result; result.up = f1.up*f2.down; result.down = f1.down*f2.up; return result; } void showResult(Fraction r){ r = reduction(r); if(r.up < 0) printf("("); if(r.down == 1) printf("%lld",r.up); else{ if(abs(r.up) > r.down){ printf("%lld %lld/%lld",r.up/r.down,abs(r.up)%r.down,r.down); }else{ printf("%lld/%lld",r.up,r.down); } } if(r.up < 0) printf(")"); } int main(){ scanf("%lld/%lld %lld/%lld",&a.up,&a.down,&b.up,&b.down); showResult(a); printf(" + "); showResult(b); printf(" = "); showResult(add(a,b)); printf("\n"); showResult(a); printf(" - "); showResult(b); printf(" = "); showResult(minu(a,b)); printf("\n"); showResult(a); printf(" * "); showResult(b); printf(" = "); showResult(multi(a,b)); printf("\n"); showResult(a); printf(" / "); showResult(b); printf(" = "); if(b.up == 0) printf("Inf"); else showResult(divide(a,b)); printf("\n"); return 0; }
标签：amp 除法 编写 pre long using code mes show
原文地址：https://www.cnblogs.com/wanghao-boke/p/10292915.html
