題目鏈接：https://www.patest.cn/contests/pat-a-PRactise/1081 Given N rational numbers in the form “numerator/denominator”, you are supposed to calculate their sum.
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.
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.
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: 2 Sample Input 3: 3 1/3 -1/6 1/8 Sample Output 3: 7/24 注意點：數據范圍為int,當兩個分母相乘時，最大可以達到long long,所以如果使用int就會溢出，有一個測試點錯誤
#include<cstdio>#include<algorithm>using namespace std;typedef long long ll;const int maxn=110;ll a[maxn],b[maxn];//分別存放分子數組，分母數組 ll gcd(ll x,ll y){//求最大公約數 if(y==0) return x; else return gcd(y,x%y);}int main(){ int n; scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%lld/%lld",&a[i],&b[i]); } ll x=b[0]; for(int i=1;i<n;i++){//求分母數組的最小公倍數 x=x*b[i]/gcd(x,b[i]); } for(int i=0;i<n;i++){ a[i]=a[i]*(x/b[i]); } ll ans1=0,ans2; for(int i=0;i<n;i++){ ans1+=a[i]; } ll t=gcd(abs(ans1),abs(x)); ans1/=t,ans2=x/t;// printf("%d %d/n",ans1,ans2); if(ans2==1){ printf("%d/n",ans1); }else if(abs(ans1)>ans2){ printf("%lld %lld/%lld/n",ans1/ans2,abs(ans1%ans2),ans2); }else{ printf("%lld/%lld/n",ans1,ans2); } return 0; }新聞熱點
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