UVaLive 7363 A Rational Sequence (二叉树)

题意:给定一个二叉树,并对每一个进行编号和规定,现在给你一个值,问你是第几个。

析:这个题,我想了好久才想出来,这个真是数据结构练的太差了,不够扎实,这个题,应该从下向上推,如果分子大于分母,那么这个编号就是奇数,

要加上1,如果是小于,就不用加.推到第一个就好。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 5;
const int mod = 1e9;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

LL f(LL a, LL b){
    if(a == b)  return 1;
    if(a < b)  return f(a, b-a) * 2;
    if(a > b)  return 2 * f(a-b, b) + 1;
}

int main(){
    int T;  cin >> T;
    while(T--){
        LL a, b;
        scanf("%d %lld/%lld", &m, &a, &b);
        printf("%d ", m);
        printf("%lld
", f(a, b));

    }
    return 0;
}

  

原文地址:https://www.cnblogs.com/dwtfukgv/p/5774465.html