# PAT 甲级 1081 Rational Sum （数据不严谨 点名批评）

2018/08/08 22:39

https://pintia.cn/problem-sets/994805342720868352/problems/994805386161274880

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代码：
#include <bits/stdc++.h>
using namespace std;

long long a[111], b[111];
long long sum, m;

long long gcd(long long x, long long y) {
long long z = x % y;
while(z) {
x = y;
y = z;
z = x % y;
}
return y;
}

long long ad(long long x, long long y) {
if(x > y)
swap(x, y);
if(y % x == 0)
return y;
else
return x * y / gcd(x, y);
}

void display(long long p, long long q) {
if(q == 0 || p == 0)
printf("0\n");
else {
bool flag = true;
if(p < 0) {
flag = false;
printf("-");
p = abs(p);
}

if(p / q != 0) {
if(p % q == 0)
printf("%lld\n", p / q);
else {
long long mm = p / q;
printf("%lld ", mm);
if(!flag) cout << "-";
printf("%lld/%lld", (p - mm * q) / gcd(p - mm * q, q), q / gcd(p - mm * q, q));
}
} else {
printf("%lld/%lld", p / gcd(p, q), q / gcd(p, q));
}

}
}

void add(long long x, long long y) {
//  sum / m + x / y
// = (sum * y + m * x) / (x * y);
long long xx = sum * y + m * x;
long long yy = m * y;
long long g = gcd(abs(xx), abs(yy));
xx /= g;
yy /= g;
sum = xx;
m = yy;
}

int main() {

int N;
scanf("%d", &N);
for(int i = 1; i <= N; i ++)
scanf("%lld/%lld", &a[i], &b[i]);

if(N == 0) {
printf("0\n");
return 0;
}
if(N ==1) {
display(a[1], b[1]);
return 0;
}

/*
long long m = ad(b[1], b[2]);
for(int i = 3; i <= N; i ++) {
m = ad(m, b[i]);
}

long long sum = 0;
for(int i = 1; i <= N; i ++) {
sum += a[i] * m / b[i];
}
*/
sum = a[1];
m = b[1];
for(int i = 2; i <= N; i ++) {
add(a[i], b[i]);
}

if(m < 0) {
sum = -sum;
m = -m;
}
display(sum, m);

return 0;
}

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