51 lines
1.2 KiB
C++
51 lines
1.2 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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int a, B, c, d;
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/**
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测试用例：
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2/1+1/3-1/4
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答案：25/12
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-2/1+1/3-1/4
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答案：-23/12
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测试点6
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7/4+1/3+8/5-3/2-6/9-6/4-5/2+6/2+4/9-3/9-2/3+5/2-5/4+3/9-3/8-5/8+6/8+3/8-4/7-5/7-3/6+6/9-5/6-5/7-5/2
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*/
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int gcd(int x, int y) {
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if (y == 0) return x;
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return gcd(y, x % y);
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}
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int lcm(int x, int y) {
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return x * y / gcd(x, y);
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}
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int main() {
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//和我一起念：scanf大法好！!
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scanf("%d/%d", &a, &b);
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//不断读取后面的数字，这玩意居然还能读入负号，牛！
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while (scanf("%d/%d", &c, &d) != EOF) {
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//通分，分母取最小公倍
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int e = lcm(b, d);
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//分子乘啊乘啊乘
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int f = e / b * a + e / d * c;
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//求最大公约数，约分
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int g = gcd(e, f);
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a = f / g;
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b = e / g;
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}
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//因为求负数最大公约，最小公倍也是可以的，但可能最终会出现1259/-360这样的情况
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//如果这时，需要特判一下，写成：-1259/360
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if (b < 0) a = -a, b = -b;
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//如果分母是1，就输出分子
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if (b == 1) printf("%d\n", a);
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else printf("%d/%d\n", a, b);
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return 0;
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} |