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简单模拟题。
#include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; struct FenShu { long long fz,fm; FenShu(long long a,long long b) { fz=a; fm=b; } }; long long gcd(long long a,long long b) { if(b==0) return a; return gcd(b,a%b); } FenShu ADD(FenShu a,FenShu b) { FenShu res(0,1); res.fz=a.fz*b.fm+b.fz*a.fm; res.fm=a.fm*b.fm; if(res.fz!=0) { long long GCD=gcd(abs(res.fz),abs(res.fm)); res.fz=res.fz/GCD; res.fm=res.fm/GCD; } else { res.fz=0; res.fm=1; } return res; } int main() { int n; scanf("%d",&n); FenShu ans(0,1); for(int i=1;i<=n;i++) { long long fz,fm; scanf("%lld/%lld",&fz,&fm); FenShu t(fz,fm); ans=ADD(ans,t); } //printf("%d/%d\n",ans.fz,ans.fm); if(ans.fz==0) printf("0\n"); else { if(ans.fz%ans.fm==0) printf("%lld\n",ans.fz/ans.fm); else if(abs(ans.fz)<ans.fm) printf("%lld/%lld\n",ans.fz,ans.fm); else { long long d=ans.fz/ans.fm; ans.fz=ans.fz-d*ans.fm; printf("%lld %lld/%lld\n",d,ans.fz,ans.fm); } } return 0; }
PAT (Advanced Level) 1081. Rational Sum (20)
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原文地址:http://www.cnblogs.com/zufezzt/p/5636195.html