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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 8/3Sample Output 1:
3 1/3Sample Input 2:
2 4/3 2/3Sample Output 2:
2Sample Input 3:
3 1/3 -1/6 1/8Sample Output 3:
7/24
测试数据比较弱。如果要算：(2^63)/(3)+(1)/(5) 怎么办？？
1 #include<cstdio> 2 #include<stack> 3 #include<algorithm> 4 #include<iostream> 5 #include<stack> 6 #include<set> 7 #include<map> 8 using namespace std; 9 long long gcd(long long a,long long b) 10 { 11 if(b==0) 12 { 13 return a; 14 } 15 return gcd(b,a%b); 16 } 17 int main() 18 { 19 //freopen("D:\\INPUT.txt","r",stdin); 20 int n,i; 21 long long fz,ffz,fm,ffm,com; 22 while(scanf("%d",&n)!=EOF) 23 { 24 scanf("%lld/%lld",&fz,&fm); 25 com=gcd(fz,fm); 26 fz/=com; 27 fm/=com; 28 for(i=1; i<n; i++) 29 { 30 //cout<<"i: "<<i<<endl; 31 scanf("%lld/%lld",&ffz,&ffm); 32 com=gcd(fm,ffm); 33 //cout<<"com: "<<com<<endl; 34 ffz=ffz*(fm/com); 35 //cout<<"ffz: "<<ffz<<endl; 36 fm=fm*(ffm/com); 37 //cout<<"fm: "<<fm<<endl; 38 fz=fz*(ffm/com); 39 //cout<<"fz: "<<ffm<<endl; 40 fz+=ffz; 41 //cout<<"fz: "<<fz<<endl; 42 com=gcd(fz,fm); 43 //cout<<"com: "<<com<<endl; 44 fz/=com; 45 //cout<<"fz: "<<fz<<endl; 46 fm/=com; 47 //cout<<"fm: "<<fm<<endl; 48 } 49 50 //cout<<fz<<" "<<fm<<endl; 51 52 if(fz%fm==0) //可以整除 53 { 54 printf("%lld\n",fz/fm); 55 } 56 else 57 { 58 if(fz/fm>1) 59 { 60 printf("%lld ",fz/fm); 61 } 62 printf("%lld/%lld\n",fz-fz/fm*fm,fm); 63 } 64 } 65 return 0; 66 }
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原文地址：http://www.cnblogs.com/Deribs4/p/4783580.html
