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(duì)分?jǐn)?shù)進(jìn)行處理，題目的關(guān)鍵在于求取最大公約數(shù)，最初我采用了循環(huán)出現(xiàn)超時(shí)，后來(lái)改用輾轉(zhuǎn)相除法，解決了此問(wèn)題。需要注意的是分子為負(fù)數(shù)的情況，為方便處理，我們把負(fù)數(shù)取絕對(duì)值，并且記錄下符號(hào)，最后再輸出。

輾轉(zhuǎn)相除法如下：

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

設(shè)a=25，b=10，c為余數(shù)

①25/10，c=5≠0，令a=10，b=5。

②10/5，c=0，則b=5就是最大公約數(shù)。

求取最大公約數(shù)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(dāng)前的最小公倍數(shù)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(zhuǎn)載于:https://www.cnblogs.com/aiwz/p/6154051.html

總結(jié)

以上是生活随笔為你收集整理的1081. Rational Sum (20) -最大公约数的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問(wèn)題。

如果覺(jué)得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。