PAT 甲级 1081 Rational Sum （数据不严谨点名批评）

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:

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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PAT 甲级 1081 Rational Sum （数据不严谨点名批评）

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巴特西

PAT 甲级 1081 Rational Sum （数据不严谨 点名批评）

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PAT 甲级 1081 Rational Sum （数据不严谨点名批评）