[PAT] A1088 Rational Arithmetic

题目大意

给出两个分数，求加减乘除的结果。

思路

求最大公约数要用辗转相除法，否则会超时。 判断符号不能两数相乘，因为乘积可能超过long long范围。

tips

题目说了输入格式一定是a1/b1 a2/b2，所以直接输入即可，不用字符串来转（在这里耗了好多时间和代码...555...我真傻，真的。）

AC代码

#define _CRT_SECURE_NO_WARNINGS #include<cstdio> #include<iostream> #include<algorithm> #include<cmath> using namespace std; int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } void huajian(long long int fenzi, long long int fenmu) { if (fenmu == 0) { printf("Inf"); return; } if (fenzi == 0) { printf("0"); return; } bool fuhao = 1; if ((fenzi < 0 && fenmu > 0) || (fenzi > 0 && fenmu < 0))fuhao = 0; if (fuhao == false)printf("(-"); if (abs(fenzi) >= abs(fenmu)) {//假分数，先求出整数部分 printf("%lld", abs(fenzi / fenmu)); fenzi = fenzi % fenmu; if (fenzi != 0)printf(" "); } if (fenzi != 0) { long long k = gcd(abs(fenzi), abs(fenmu)); fenzi = fenzi / k;fenmu = fenmu / k; printf("%lld/%lld", abs(fenzi), abs(fenmu)); } if (fuhao == false)printf(")"); } int main() { long long int a1, b1, a2, b2; scanf("%lld/%lld %lld/%lld", &a1, &b1, &a2, &b2); huajian(a1, b1);printf(" + ");huajian(a2, b2);printf(" = "); huajian(a1 * b2 + a2 * b1, b1 * b2);printf("\n"); huajian(a1, b1);printf(" - ");huajian(a2, b2);printf(" = "); huajian(a1 * b2 - a2 * b1, b1 * b2);printf("\n"); huajian(a1, b1);printf(" * ");huajian(a2, b2);printf(" = "); huajian(a1 * a2, b1 * b2);printf("\n"); huajian(a1, b1);printf(" / ");huajian(a2, b2);printf(" = "); huajian(a1 * b2, b1 * a2);printf("\n"); return 0; }

傻傻的代码，删了可惜QAQ...

#define _CRT_SECURE_NO_WARNINGS #include<cstdio> #include<iostream> #include<string> #include<map> #include<algorithm> #include<cmath> using namespace std; struct node { long long int fenzi, fenmu; node() { fenzi = 0; fenmu = 0; } }; int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } void huajian(long long int fenzi, long long int fenmu) { if (fenmu == 0) { printf("Inf"); return; } if (fenzi == 0) { printf("0"); return; } bool fuhao = 1; if ((fenzi < 0 && fenmu > 0) || (fenzi > 0 && fenmu < 0))fuhao = 0; if (fuhao == false)printf("(-"); if (abs(fenzi) >= abs(fenmu)) {//假分数，先求出整数部分 printf("%lld", abs(fenzi / fenmu)); fenzi = fenzi % fenmu; if (fenzi != 0)printf(" "); } if (fenzi != 0) { long long k = gcd(abs(fenzi), abs(fenmu)); fenzi = fenzi / k;fenmu = fenmu / k; printf("%lld/%lld", abs(fenzi), abs(fenmu)); } if (fuhao == false)printf(")"); } int main() { node op[2]; for (int i = 0;i < 2;i++) { string stemp; int flag = true;//负数是0 cin >> stemp; if (stemp[0] == '-') { flag = false; stemp.erase(stemp.begin()); } int j = 0; while (stemp[j] != '/' && j < stemp.size()) { op[i].fenzi = op[i].fenzi * 10 + stemp[j] - '0'; j++; } if (flag == false)op[i].fenzi = -op[i].fenzi; if (j == stemp.size())op[i].fenmu = 1; j++; while (j < stemp.size()) { op[i].fenmu = op[i].fenmu * 10 + stemp[j] - '0'; j++; } } huajian(op[0].fenzi, op[0].fenmu);printf(" + ");huajian(op[1].fenzi, op[1].fenmu);printf(" = "); huajian(op[0].fenzi * op[1].fenmu + op[1].fenzi * op[0].fenmu, op[0].fenmu * op[1].fenmu);printf("\n"); huajian(op[0].fenzi, op[0].fenmu);printf(" - ");huajian(op[1].fenzi, op[1].fenmu);printf(" = "); huajian(op[0].fenzi * op[1].fenmu - op[1].fenzi * op[0].fenmu, op[0].fenmu * op[1].fenmu);printf("\n"); huajian(op[0].fenzi, op[0].fenmu);printf(" * ");huajian(op[1].fenzi, op[1].fenmu);printf(" = "); huajian(op[0].fenzi * op[1].fenzi, op[0].fenmu * op[1].fenmu);printf("\n"); huajian(op[0].fenzi, op[0].fenmu);printf(" / ");huajian(op[1].fenzi, op[1].fenmu);printf(" = "); huajian(op[0].fenzi * op[1].fenmu, op[0].fenmu * op[1].fenzi);printf("\n"); return 0; }