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[],a2[];
 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-){
         A = arr[cur];
         B = ;
     }
     if(cur >= dep-) return;
     LL tmpA,tmpB;
     dfs(tmpA,tmpB,arr,cur+,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?' ':'\n');
     if(p) print(y,p);
 }
 int main(){
     while(~scanf("%d %d",&n,&m)){
         for(int i = ; i < n; ++i) scanf("%d",a1+i);
         for(int i = ; i < m; ++i) scanf("%d",a2+i);
         LL A1 = ,B1 = ,A2 = ,B2 = ;
         dfs(B1,A1,a1,,n);
         dfs(B2,A2,a2,,m);
         A1 += a1[]*B1;
         A2 += a2[]*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 ;
 }