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宇宙第一小仙女\(^o^)/～～萌量爆表求帶飛=≡Σ((( つ^o^)つ~ dalao們點個關注唄～～
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一、問題：
Description
An infinite full binary tree labeled by positive rational numbers is defined by:
• The label of the root is 1/1.
• The left child of label p/q is p/(p+q).
• The right child oflabel p/q is (p+q)/q.
The top of the tree is shown in the following figure:
A rational sequence is defined by doing a level order (breadth first) traversal of the tree (indicated
by the light dashed line). So that:
F(1) = 1/1, F(2) = 1/2, F(3) = 2/1, F(4) = 1/3, F(5) = 3/2, F(6) = 2/3, . . .
Write a program which takes as input a rational number, p/q, in lowest terms and finds the next
rational number in the sequence.
That is, if F(n) = p/q, then the result is F(n + 1).
Input
The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets
that follow. Each data set should be processed identically and independently.
Each data set consists of a single line of input. It contains the data set number, K, which is then
followed by a space, then the numerator of the fraction, p, followed immediately by a fonward slash (/),
followed immediately by the denominator of the fraction, q. Both p and q will be relatively prime and
0 ≤ p, q ≤ 2147483647.
Output
For each data set there is a single line of output. It contains the data set number, K, followed by a
single space which is then followed by the numerator of the fraction, followed immediately by a forward
slash (‘/’) followed immediately by the denominator of the fraction. Inputs will be chosen such that
neither the numerator nor the denominator will overflow a 32-bit integer.
Sample Input
5
1 1/1
2 1/3
3 5/2
4 2178309/1346269
5 1/10000000
Sample Output
1 1/2
2 3/2
3 2/5
4 1346269/1860498
5 10000000/9999999
二、題意：
如圖所示，根爲1/1，假設某結點是p/q，則其左子樹爲p/(p+q)，右子樹爲(p+q)/q，按此規律畫一棵樹，已知第n個結點是p/q，求第n+1個結點。
三、思路：
其實之前做過一個非常類似的題目，戳這裏～，那個題是給出n求第n個結點，受那題影響，於是做這個題爲最初的思路是：除了特判，其他的由當前結點開始，向上搜，每次記錄，用0和1分別表示當前結點是上一結點的左子樹還是右子樹，然後再從根結點回來找到下一個，很不幸，RE了5發，按此思路也能做，但是很麻煩，易出錯，沒必要。
在最後還剩半個小時的時候，想出了第二種思路，簡單的不能再簡單了，但是來不及敲了。下面介紹一下第二種思路，畫出前五層，仔細觀察，就發現規律了，無非就以下5種情況：
1）根結點，1/1，注意它的下一結點是1/2不是2/1。
2）p=1，即樹最左邊的那一列；q=1，即樹最右邊那一列。
3）q=2，即樹最中心的兩列，位置對稱，值也對稱。
4）p<q，即左子樹，推其父親節點爲a/(b-a)，由此再算右子樹即可。
5）p>q，即右子樹，此種情況最爲麻煩，由當前結點不斷找父親節點，直到找到第一個p<q（是左子樹）的父親節點A，記錄下尋找次數ans，然後順着該結點的兄弟B的左子樹找ans次，順推出結果。注意此時無需考慮下一個結點會是樹的下一層，此情況已包含在第2種情況中。
四、代碼：
#include<cstdio>
int main()
{
int T,k;
long long p,q;
scanf("%d",&T);
while(T--)
{
scanf("%d %lld/%lld",&k,&q,&p);
if(p==1 && q==1)
printf("%d 1/2\n",k);
else if(p==1)
printf("%d 1/%lld\n",k,q+1);
else if(q==1)
printf("%d %lld/%lld\n",k,p,p-q);
else if(p==2)
printf("%d %lld/%lld\n",k,p,q);
else if(q<p)
printf("%d %lld/%lld\n",k,p,p-q);
else
{
int ans=0;
while(p<=q)
{
q-=p;
ans++;
}
printf("%d %lld/%lld\n",k,p,p-q+ans*p);
}
}
return 0;
}