CSUOJ 1638 Continued Fraction

1638: Continued Fraction

Time Limit: 1 Sec  Memory Limit: 128 MB

Description

 

 

 

 

 

 

 

Input

Output

Sample Input

4 3
5 1 1 2
5 2 2

Sample Output

11
0 5
30 4 6
1 27

HINT

 

Source

解题：主要任务是把任一一个分数化成连分式。

 

方法就是分子分母同时不断的除以分子，直到分子为0。

 

至于加，减，乘，除都是先算出分数，然后把分数化成连分数。。

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int n,m,a1[20],a2[20];
LL gcd(LL x,LL y){
    return y?gcd(y,x%y):x;
}
void dfs(LL &A,LL &B,int *arr,int cur,int dep){
    if(cur == dep-1){
        A = arr[cur];
        B = 1;
    }
    if(cur >= dep-1) return;
    LL tmpA,tmpB;
    dfs(tmpA,tmpB,arr,cur+1,dep);
    LL theGCD = gcd(A = arr[cur]*tmpA + tmpB,B = tmpA);
    A /= theGCD;
    B /= theGCD;
}
void print(LL x,LL y){
    LL GCD = gcd(x,y);
    LL tmp = (x /= GCD)/(y /= GCD),p = x - tmp*y;
    printf("%lld%c",tmp,p?' ':'
');
    if(p) print(y,p);
}
int main(){
    while(~scanf("%d %d",&n,&m)){
        for(int i = 0; i < n; ++i) scanf("%d",a1+i);
        for(int i = 0; i < m; ++i) scanf("%d",a2+i);
        LL A1 = 0,B1 = 1,A2 = 0,B2 = 1;
        dfs(B1,A1,a1,1,n);
        dfs(B2,A2,a2,1,m);
        A1 += a1[0]*B1;
        A2 += a2[0]*B2;
        print(A1*B2 + A2*B1,B1*B2);
        print(A1*B2 - A2*B1,B1*B2);
        print(A1*A2,B1*B2);
        print(A1*B2,A2*B1);
    }
    return 0;
}