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。。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 int n,m,a1[20],a2[20];
 5 LL gcd(LL x,LL y){
 6     return y?gcd(y,x%y):x;
 7 }
 8 void dfs(LL &A,LL &B,int *arr,int cur,int dep){
 9     if(cur == dep-1){
10         A = arr[cur];
11         B = 1;
12     }
13     if(cur >= dep-1) return;
14     LL tmpA,tmpB;
15     dfs(tmpA,tmpB,arr,cur+1,dep);
16     LL theGCD = gcd(A = arr[cur]*tmpA + tmpB,B = tmpA);
17     A /= theGCD;
18     B /= theGCD;
19 }
20 void print(LL x,LL y){
21     LL GCD = gcd(x,y);
22     LL tmp = (x /= GCD)/(y /= GCD),p = x - tmp*y;
23     printf("%lld%c",tmp,p?‘ ‘:‘\n‘);
24     if(p) print(y,p);
25 }
26 int main(){
27     while(~scanf("%d %d",&n,&m)){
28         for(int i = 0; i < n; ++i) scanf("%d",a1+i);
29         for(int i = 0; i < m; ++i) scanf("%d",a2+i);
30         LL A1 = 0,B1 = 1,A2 = 0,B2 = 1;
31         dfs(B1,A1,a1,1,n);
32         dfs(B2,A2,a2,1,m);
33         A1 += a1[0]*B1;
34         A2 += a2[0]*B2;
35         print(A1*B2 + A2*B1,B1*B2);
36         print(A1*B2 - A2*B1,B1*B2);
37         print(A1*A2,B1*B2);
38         print(A1*B2,A2*B1);
39     }
40     return 0;
41 }