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PAT:B1034 有理数四则运算
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
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)
输入样例 2:
5/3 0/6
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
代码:
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
// 本题用到了分数的加减,化简,寻找公约数, 分子的输出
int gcd(ll a, ll b) {
if(b == 0) return a;
else return gcd(b, a%b);
}
// 分数的构造体
struct Fraction{
long long up, down;
};
// 分数的化简
Fraction reduction(Fraction r) {
if(r.down < 0) {
r.up = -r.up;
r.down = -r.down;
}
if(r.up == 0) r.down = 1;
else {
ll d = gcd(abs(r.up), abs(r.down));
r.up = r.up / d;
r.down = r.down / d;
}
return r;
}
// 分数的加法
Fraction add(Fraction a, Fraction b) {
a = reduction(a), b = reduction(b);
Fraction c;
c.down = a.down * b.down;
c.up = a.up * b.down + a.down * b.up;
return reduction(c);
}
// 分数的减法
Fraction sub(Fraction a, Fraction b) {
a = reduction(a), b = reduction(b);
Fraction c;
c.down = a.down * b.down;
c.up = a.up * b.down - a.down * b.up;
return reduction(c);
}
// 分子的乘法
Fraction mult(Fraction a, Fraction b) {
Fraction c;
c.up = a.up * b.up;
c.down = a.down * b.down;
return reduction(c);
}
// 分子的除法
Fraction divide(Fraction a, Fraction b) {
Fraction c;
c.up = a.up * b.down;
c.down = a.down * b.up;
return reduction(c);
}
// 分数的输出
void printFra(Fraction r) {
r = reduction(r);
if(r.up * r.down < 0) {
if(r.down == 1) printf("(%lld)", r.up);
else if(abs(r.up) > abs(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);
}
} else {
if(r.up == 0) printf("%lld", r.up);
else if(r.down == 1) printf("%lld", r.up);
else if(r.up > r.down) printf("%lld %lld/%lld", r.up/r.down, r.up%r.down, r.down);
else {
printf("%lld/%lld", r.up, r.down);
}
}
}
int main() {
Fraction a, b;
scanf("%lld/%lld %lld/%lld", &a.up, &a.down, &b.up, &b.down);
char alg[4][5] = {" + ", " - ", " * ", " / "};
for(int i = 0; i < 4; i++) {
printFra(a);
printf("%s", alg[i]);
printFra(b);
printf(" = ");
switch(i) {
case 0:
printFra(add(a, b));
break;
case 1:
printFra(sub(a, b));
break;
case 2:
printFra(mult(a, b));
break;
default:
if(b.up == 0) {
printf("Inf");
} else {
printFra(divide(a, b));
}
}
if(i != 3) printf("\n");
}
return 0;
}