1081 Rational Sum (20)
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:
7/2
4 题目大意：给N个有理数（以分子/分母的形式给出），计算这N个数的总和，末了总和要以（整数 分子/分母）的形式给出。如果整数部分为0，则直接输出小数部分。
分析：用分数加法举行模仿，每次先把分母变成两个分数的最小公倍数，再把分子乘以相应倍数相加。
/*********************************************** * _ooOoo_ * * o8888888o * * 88" . "88 * * (| -_- |) * * O\=/O * * ____/`---'\____ * * .'\\| |//`. * * /\\|||:|||//\ * * /_||||| -:- |||||-\ * * | | \\\- * | | * * | \_|''\---/''| | * * \.-\__`-`___/-. / * * ___`. .'/--.--\`. . __ * * ."" '<`.___\_<|>_/___.'>'"". * * | | :`- \`.;`\ _ /`;.`/ - ` : | | * * \\ `-. \_ __\ /__ _/ .-` // * *======`-.____`-.___\_____/___.-`____.-'======* * `=---=' * *^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 佛祖保佑 永无BUG 本程序已经经过开光处理，绝无可能再产生bug **********************************************/#include<algorithm>#include <iostream>#include<cstdlib>#include<cstring>#include <string>#include <vector>#include <cstdio>#include <queue>#include <stack>#include <ctime>#include <cmath>#include <map>#include <set>#define INF 0xffffffff#define db1(x) cout<<#x<<"="<<(x)<<endl#define db2
(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endlusing namespace std;long long gcd(long long a,long long b){ return b==0?a:gcd(b,a%b);}long long lcm(long long a,long long b){ return a/gcd(a,b)*b;}int main(void){ #ifdef test freopen("in.txt","r",stdin); //freopen("in.txt","w",stdout); clock_t start=clock(); #endif //test int n;scanf("%d",&n); long long numerator=0,denominator=1; for(int i=0;i<n;++i) { long long a,b;scanf("%lld/%lld",&a,&b); long long t1,t2
,temp=lcm(denominator,b); t1=temp/denominator,t2
=temp/b; denominator=temp; numerator=numerator*t1+a*t2
; long long g=gcd((long long)fabs(denominator),(long long)fabs(numerator)); denominator/=g,numerator/=g; } int f=0;// db2
(denominator,numerator); if(numerator%denominator==0)printf("%lld\n",numerator/denominator); else { long long x=numerator/denominator;// db3(x,numerator,denominator); numerator=numerator-x*denominator; if(x!=0)printf("%lld ",x); if(denominator<0) { if(numerator<0)printf("%lld/%lld\n",-1*numerator,-1*denominator); else printf("-%lld/%lld\n",numerator,-1*denominator); } else printf("%lld/%lld\n",numerator,denominator); } #ifdef test clockid_t end=clock(); double endtime=(double)(end-start)/CLOCKS_PER_SEC; printf("\n\n\n\n\n"); cout<<"Total time:"<<endtime<<"s"<<endl; //s为单位 cout<<"Total time:"<<endtime*1000<<"ms"<<endl; //ms为单位 #endif //test return 0;}// ━━━━━━神兽出没━━━━━━// ┏┓ ┏┓// ┏┛┻━━━━━━━┛┻┓// ┃ ┃// ┃ ━ ┃// ????━???? ┃// ┃ ┃// ┃ ┻ ┃// ┃ ┃// ┗━┓ ┏━┛// ┃ ┃// ┃ ┃// ┃ ┗━━━┓// ┃ ┣┓// ┃ ┏┛// ┗┓┓┏━━━━━┳┓┏┛// ┃┫┫ ┃┫┫// ┗┻┛ ┗┻┛//// ━━━━━━感觉萌萌哒━━━━━━
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