PAT (Advanced Level) Practise 1081 Rational Sum (20)

1081. Rational Sum (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

52/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

24/3 2/3

Sample Output 2:

2

Sample Input 3:

31/3 -1/6 1/8

Sample Output 3:

7/24

分数加减，注意判断特殊情况。

#include<cstdio>#include<vector>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int maxn = 1e5 + 10;LL n, x, y, z, a, b, c;char s[maxn];LL gcd(LL x, LL y){if (!x || !y) return x | y;return x%y ? gcd(y, x%y) : y;}void add(){if (c == 0) { a = x, b = y, c = z; return; }LL g = gcd(c, z);LL gg = c / g * z;b = z / g * b;y = c / g * y;if (a == x) b += y;else if (a == 0){if (b >= y) b -= y; else a = 1, b = y - b;}else{if (b > y) b -= y; else a = 0, b = y - b;}g = gcd(gg, b);b /= g;c = gg / g;}int main(){scanf("%lld", &n);while (n--){scanf("%s", s);x = s[0] == '-';sscanf(s + x, "%lld/%lld", &y, &z);LL g = gcd(y, z);y /= g;z /= g;add();}if (a) printf("-");if (b / c) printf("%lld", b / c);if (b / c && b%c) printf(" ");if (b % c) printf("%lld/%lld", b%c, c);if (!(b / c) && !(b%c)) printf("0");return 0;}