POJ 2499 Binary Tree

题意：二叉树的根节点为(1,1)，对每个结点(a,b)其左结点为 (a + b, b) ，其右结点为 (a, a + b)，已知某结点坐标，求根节点到该节点向左和向右走的次数。

分析：往回一步一步走肯定超时，不过如果a>b，大的有点多，就没必要一步一步走，直接a/b就好了，然后把a变成a%b取余，那么a/b就是从现在的a到原来的a向左走的步数。(a<b同理)

同理，如果a=1的话，那么也没必要往回走了，因为从根节点到现在的结点就是向右走了b-1。(b=1同理)

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1};
const int dc[] = {-1, 1, 0, 0};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 10000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int main(){
    int T;
    scanf("%d", &T);
    for(int i = 1; i <= T; ++i){
        printf("Scenario #%d:\n", i);
        int a, b;
        scanf("%d%d", &a, &b);
        int l, r;
        l = r = 0;
        while(1){
            if(a == 1){
                r += b - 1;
                break;
            }
            if(b == 1){
                l += a - 1;
                break;
            }
            if(a > b){
                l += a / b;
                a = a % b;
            }
            else if(a < b){
                r += b / a;
                b = b % a;
            }
        }
        printf("%d %d\n", l, r);
        printf("\n");
    }
    return 0;
}