[PAT解题报告] Rational Sum

#include <bits/stdc++.h> using namespace std; typedef long long LL; struct Node {     int fz,fm; }divs[1010]; LL GCD(LL a,LL b) {     return b==0?a:GCD(b,a%b); } Node mul(Node a,Node b) {     Node ans;     int tfz = a.fz*b.fm+a.fm*b.fz;     int tfm = a.fm*b.fm;     //cout<<tfz<<" "<<tfm<<endl;     int div = abs(GCD(tfz,tfm));    // cout<<div<<endl;     tfz/=div;     tfm/=div;     ans.fm = tfm;     ans.fz = tfz;     return ans; } void strcin(string s,Node &T) {     int len = s.length();     int ans = 0;     int Rval = 1;     for(int i=0;i<len;i++)     {         if(s[i]=='-') Rval  = -1;         else if(s[i]=='/')            T.fz = ans*Rval,ans=0;         else ans=ans*10+(s[i]-'0');     }     T.fm = ans;     return; } int main() {     string s;     int T;     int index = 0;     scanf("%d",&T);     while(T--)     {         cin>>s;         strcin(s,divs[index++]);     }     Node ans = divs[0];     for(int i=1;i<index;i++)         ans = mul(ans,divs[i]);    // cout<<ans.fz<<" "<<ans.fm<<endl;    int Inpart = ans.fz/ans.fm;    //cout<<Inpart<<endl;    ans.fz%=ans.fm;    int Fapart = ans.fz;    if(Inpart!=0)    {        cout<<Inpart;        if(Fapart==0) cout<<endl;        else cout<<" "<<Fapart<<"/"<<ans.fm<<endl;    }    else    {        if(Fapart==0) cout<<0<<endl;        else         cout<<Fapart<<"/"<<ans.fm<<endl;    }     return 0; }

package go.jacob.day822; import java.util.Scanner; public class Demo1 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); String[] strs = new String[n]; //三个数分别为分子，分母，带分数的整数部分 int numerator = 0, denominator = 0, first = 0; for (int i = 0; i < n; i++) strs[i] = sc.next(); sc.close(); for (int i = 0; i < n; i++) { int flag = 1, begin = 0; if (strs[i].charAt(0) == '-') { flag = -1; begin = 1; } int tmpNumerator = flag * Integer.parseInt(strs[i].substring(begin, strs[i].indexOf("/"))); int tmpDenominator = Integer.parseInt(strs[i].substring(strs[i].indexOf("/") + 1)); if (tmpDenominator == 0) return; if (i == 0) { numerator = tmpNumerator; denominator = tmpDenominator; } else { numerator = numerator * tmpDenominator + tmpNumerator * denominator; denominator *= tmpDenominator; } // 使用欧几里得算法求最大公约数，每次计算都要对分数进行化简，放置溢出 int factor = gcd(numerator, denominator); if (factor == 0) { System.out.println(0); return; } numerator /= factor; denominator /= factor; } // 把假分数化成带分数 first += numerator / denominator; numerator = numerator % denominator; // 输出结果的时候要判断：整数部分是否为0？分子书否为0？ if (first != 0) if (numerator != 0) System.out.println(first + " " + numerator + "/" + denominator); else System.out.println(first); else if (numerator != 0) System.out.println(numerator + "/" + denominator); else System.out.println(0); } private static int gcd(int a, int b) { if (a == 0) return 0; if (a < 0) a = -a; if (b < 0) b = -b; if (a < b) { int tmp = a; a = b; b = tmp; } if (a % b == 0) return b; else return gcd(b, a % b); } }

示例输出应该是 10/3