UVaOJ 694 - The Collatz Sequence】的更多相关文章

题目很简单,但是一开始却得到了Time Limit的结果,让人感到很诧异.仔细阅读发现,题目中有一个说明: Neither of these, A or L, is larger than 2,147,483,647 (the largest value that can be stored in a 32-bit signed integer). 所以应该是A溢出造成了程序“死循环”,最终导致超时. -----------------------------------------------…
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=94&page=show_problem&problem=635 题目:给出了如下算法: step 1:选择任意一个正整数A为序列的第一个数: step 2:如A=1或者A>L,则停止 step 3:如A为偶数,则A=A/2,执行step 2: step 4:如A为奇数,则A=3A+1,执行step 2: 现在给…
The following iterative sequence is defined for the set of positive integers: n n/2 (n is even) n 3n + 1 (n is odd) Using the rule above and starting with 13, we generate the following sequence: 13 40 20 10 5 16 8 4 2 1 It can be seen that this seque…
title: The following iterative sequence is defined for the set of positive integers: n n/2 (n is even) n 3n + 1 (n is odd) Using the rule above and starting with 13, we generate the following sequence: 13 40 20 10 5 16 8 4 2 1 It can be seen that thi…
题意:对于任意一个数 N ,寻找在 100,0000 之内按照规则( N 为奇数 N = N * 3 + 1 ,N 为偶数 N = N / 2 ,直到 N = 1 时的步数 )步数的最大值 思路:记忆化搜索即可,利用之前搜索的值加速搜索,如果当前搜索值在之前已经处理过,那么直接利用当前搜索值 + 到当前数的步数即为该数的步数 /************************************************************************* > File Name…
Language: C++ 4.8.2 #include<stdio.h> int main(void) { long long int m, n, copy_m; ; ; ) { scanf("%lld%lld", &m, &n); // uva要求用%lld读入有符号长整型 && n < ) break; count = ; copy_m = m; ) { == ) m = m / ; else { m = *m + ; if(m…
记忆化搜索来一发.没想到中间会爆int #include <bits/stdc++.h> using namespace std; const int MAXN = 1000000; int dp[MAXN]; int dfs(long long num) { if(num <= MAXN && dp[num]) return dp[num]; if(num == 1) return (dp[num] = 1); if(num&1) { if(num <=…
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