C++题解1081 Ration

Given N rational numbers in the form numerator/denominator, you are supposed to calculate their sum.

给定N个有理数，分子/分母，要求你计算它们的和

Input Specification:
Each input file contains one test case. Each case starts with a positive integer N (≤100), followed in the next line N rational numbers a1/b1 a2/b2 … where all the numerators and denominators are in the range of long int. If there is a negative number, then the sign must appear in front of the numerator.

输入规格：每个输入文件包含一个测试建立，每个测试案例，第一个是N，紧接着N个有理数据，a1/b1,a2/b2,…所有的分子和分母都是长整型，如果是负数，就必须出现在分子的前面

Output Specification:
For each test case, output the sum in the simplest form integer numerator/denominator where integer is the integer part of the sum, numerator < denominator, and the numerator and the denominator have no common factor. You must output only the fractional part if the integer part is 0.

针对每个测试案例，输出他们的和，数字,分子/分母数字时候和的整数部分，分子<分母比你高且分子和分母没有相同的因子，如果整数部分为0那就输出小数部分。

Sample Input 1:

5
2/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

2
4/3 2/3

Sample Output 2:

Sample Input 3:

3
1/3 -1/6 1/8

Sample Output 3:

7/24

思路核心

学会最大公约数，最小公倍数，模拟出分数加和通分的过程。

完整代码

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll; //记住ll为long long
int gcd(ll a,ll b){//求a和b最大公约数if(b==0) return a;else return gcd(b,a%b);
}
struct Fraction{ll up,down;
};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 f1,Fraction f2){Fraction res;res.up = f1.up*f2.down + f2.up*f1.down;res.down = f1.down * f2.down;return reduction(res);
}
void showResult(Fraction r){reduction(r);if(r.down == 1) printf("%lld\n",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\n",r.up,r.down);}
}
int main()
{//freopen("C:\\Users\\Administrator\\Desktop\\test\\input.txt","r",stdin);Fraction res{0,1};int n;scanf("%d",&n);for(int i =0;i<n;i++){Fraction tmp;scanf("%lld/%lld",&tmp.up,&tmp.down);res = add(res,tmp);}showResult(res);return 0;
}

方法2，手动模拟分数

#include<iostream>
using namespace std;
typedef long long ll;
ll a[100],b[100],sigmal[100];ll gy(ll x,ll y){ll big,small;if(x>y){big = x;small = y;}else{big = y;small = x;}while(small!=0){long tmp = big%small;big = small;small = tmp;}return big;
}ll gbs(ll x,ll y){ll tmp = gy(x,y);return x*(y/tmp);
}int main()
{// freopen("C:\\Users\\Administrator\\Desktop\\test\\input.txt","r",stdin);ll i,j,k,l,N;scanf("%lld",&N);for(i = 0;i<N;i++){scanf("%lld/%lld",&a[i],&b[i]);if(a[i]<0){a[i]*=-1;sigmal[i]=-1;}elsesigmal[i] = 1;}ll A,B;B = b[0];for(i = 1;i<N;i++){B = gbs(B,b[i]);}A= 0;for(i = 0;i<N;i++){A+= sigmal[i] * a[i]*(B/b[i]);}//分子累加ll tmp = gy(A,B);A/=tmp;B/=tmp;if(A==0){cout << 0;return 0;}if(A<0){cout << "-";A*=-1;}if(A>=B){cout << A/B;}if(A>=B&&A%B){cout << " ";}if(A%B){cout << A%B << "/" << B;}return 0;
}