Rational Sum
题意:用程序模拟分数的加法。
首先定义一个结构体来存储分子,分母。
因为分子分母都是int类型,相乘有可能溢出,所以用long long 类型保存。
typedef long long ll;
struct Fraction
{
ll up, down;
};
分子分母表示时必须是最简的,所以分子分母要同时除以最大公约数。
最大公约数:假设有两个int类型的数据a,b(a > b), 而a = kb + r;设d为a,b的公约数,而r = kb - a,所以d也为r的公约数。而
r = a % b,所以d也为b,a%b的公约数。而寻找a, b的公约数可以近一步缩小为b,a%b的公约数。当a%b == 0时,说明a = nb,所以此时b是最大公约数。
所以最大公约数求法:
int gcd(ll a, ll b)
{
if(b == 0)
return a;
else
return gcd(b, a%b);
}
EXTRA:最大公倍数:ab/d,有时ab会溢出,可以用a/db;
分数的化简:
Fraction reduction(Fraction result){
if(result.down < 0)
{
result.up = -result.up;
result.down = -result.down;
}//分母小于0时,分子取相反数,分母变为正数好输出
if(result.up == 0)
result.down = 1;
else
{
int d = gcd(abs(result.up), abs(result.down));
result.up /= d;
result.down /= d;
}
return result;
}
分数的加减法:
Fraction add(Fraction a, Fraction b)
{
Fraction result;
result.up = a.up * b.down + b.up * a.down;
//减法这里+号改为减号
result.down = a.down * b.down;
return reduction(result);
}
乘法:
Fraction mul(Fraction a, Fraction b)
{
Fraction result;
result.up = a.up * b.up;
result.down = a.down * b.down;
return reduction(result);
}
除法:
Fraction div(Fraction a, Fraction b)
{
Fraction result;
result.up = a.up * b.down;
result.down = a.down * b.up;
return reduction(result);
}
输出:
void showResult(Fraction a)
{
reduction(a);//注意这里要化简
if(a.down == 1)
{
printf("%lld\n", a.up);
}
else if(abs(a.up) > a.down)
{
printf("%lld %lld/%lld\n", a.up/a.down, abs(a.up) % a.down, a.down);
}
else
printf("%lld/%lld\n", a.up, a.down);
}
完整程序:
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
struct Fraction
{
ll up, down;
};
int gcd(ll a, ll b)
{
return !b ? a : gcd(b, a % b);
}
Fraction reduction(Fraction result){
if(result.down < 0)
{
result.up = -result.up;
result.down = -result.down;
}
if(result.up == 0)
result.down = 1;
else
{
int d = gcd(abs(result.up), abs(result.down));
result.up /= d;
result.down /= d;
}
return result;
}
Fraction add(Fraction a, Fraction b)
{
Fraction result;
result.up = a.up * b.down + b.up * a.down;
result.down = a.down * b.down;
return reduction(result);
}
void showResult(Fraction a)
{
reduction(a);
if(a.down == 1)
{
printf("%lld\n", a.up);
}
else if(abs(a.up) > a.down)
{
printf("%lld %lld/%lld\n", a.up/a.down, abs(a.up) % a.down, a.down);
}
else
printf("%lld/%lld\n", a.up, a.down);
}
int main()
{
int n;
Fraction sum, temp;
sum.up = 0;
sum.down = 1;
scanf("%d", &n);
for(int i = 0; i < n; i++)
{
scanf("%lld/%lld", &temp.up, &temp.down);
sum = add(sum, temp);
}
showResult(sum);
return 0;
}