題目如下：

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

題目要求對分數進行處理，題目的關鍵在于求取最大公約數，最初我采用了循環出現超時，后來改用輾轉相除法，解決了此問題。需要注意的是分子為負數的情況，為方便處理，我們把負數取絕對值，并且記錄下符號，最后再輸出。

輾轉相除法如下：

給定數a、b，要求他們的最大公約數，用任意一個除以另一個，得到余數c，如果c=0，則說明除盡，除數就是最大公約數；如果c≠0，則用除數再去除以余數，如此循環下去，直至c=0，則除數就是最大公約數，直接說比較抽象，下面用例子說明。

設a=25，b=10，c為余數

①25/10，c=5≠0，令a=10，b=5。

②10/5，c=0，則b=5就是最大公約數。

求取最大公約數的代碼如下：

long getMaxCommon(long a, long b){long yu;if(a == b) return a;while(1){yu = a % b;if(yu == 0) return b;a = b;b = yu;} }

完整代碼如下：

#include <iostream> #include <stdio.h> #include <vector>using namespace std;struct Ration{long num;long den;Ration(long _n, long _d){num = _n;den = _d;}};long getMaxCommon(long a, long b){long yu;if(a == b) return a;while(1){yu = a % b;if(yu == 0) return b;a = b;b = yu;} }int main(){int N;long num,den;long maxDen = -1;cin >> N;vector<Ration> rations;for(int i = 0; i < N; i++){scanf("%ld/%ld",&num,&den);rations.push_back(Ration(num,den));if(maxDen == -1){maxDen = den;}else{// 找maxDen和當前的最小公倍數if(den == maxDen) continue;else if(maxDen > den){if(maxDen % den == 0) continue;}else{if(den % maxDen == 0){maxDen = den;continue;}}maxDen = maxDen * den;}}num = 0;for(int i = 0; i < N; i++){num += rations[i].num * (maxDen / rations[i].den);}if(num == 0) {printf("0\n");return 0;}bool negative = num < 0;if(negative) num = -num;if(num >= maxDen){long integer = num / maxDen;long numerator = num % maxDen;if(numerator == 0){if(negative)printf("-%ld\n",integer);elseprintf("%ld\n",integer);return 0;}long common = getMaxCommon(numerator,maxDen);if(negative){printf("%ld -%ld/%ld\n",integer,numerator/common,maxDen / common);}else{printf("%ld %ld/%ld\n",integer,numerator/common,maxDen / common);}}else{long common = getMaxCommon(num,maxDen);if(negative)printf("-%ld/%ld\n",num/common,maxDen/common);elseprintf("%ld/%ld\n",num/common,maxDen/common);}return 0; }

轉載于:https://www.cnblogs.com/aiwz/p/6154051.html

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