说说:此题要求求出从整数x到达整数y所要经过的最短步数,且第一步和最后一步必须为一,同一时候每一步都比前一步多一步,少一步或一样.如果想搞清楚每一步详细是如何走的,那么这道题是相当麻烦的.考虑到前后两步之间最多差一,那么一到最大数之间的每个数从左到右以及从右到左的时候必然都出现.那么我们能够预想如果这样一个数列. 1,2,3,....n-2,n-1,n,n-1,n-2....3,2,1,若数列的和小于y和x的差肯定是不行的.所以要调整n至刚好大于或等于y和x的差.显然当y与x的差恰好为数列的和那…

找出步數與距離的關係即可得解. 0步最多能抵達的距離是0 1步最多能抵達的距離是1(1) 2步最多能抵達的距離是2(1 1) 3步最多能抵達的距離是4(1 2 1) 4步最多能抵達的距離是6(1 2 2 1) 5步最多能抵達的距離是9(1 2 3 2 1) 6步最多能抵達的距離是12(1 2 3 3 2 1) ……以此類推 #include <iostream> #include <cstdio> #include <cmath> #define ERROR 1e-10…

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

// The 3n+1 problem (3n+1 问题) // PC/UVa IDs: 110101/100, Popularity: A, Success rate: low Level: 1 // Verdict: Accepted // Submission Date: 2011-05-22 // UVa Run Time: 0.032s // // 版权所有(C)2011,邱秋.metaphysis # yeah dot net. // // [问题描述] // 考虑如下的序列生成算法…

, '\n'); #include<cstdio> #include<iostream> #include<string> #include<algorithm> #include<iterator> #include<cstring> //uva 10033 Problem G: Interpreter #define ONLINE_JUDGE using namespace std; ]; ]; int steps; ; int…

uva 116 Unidirectional TSP Background Problems that require minimum paths through some domain appear in many different areas of computer science. For example, one of the constraints in VLSI routing problems is minimizing wire length. The Traveling Sa…

UVA - 1153 Keep the Customer Satisfied Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Description   Simon and Garfunkel Corporation (SG Corp.) is a large steel-making company with thousand of customers. Keeping the customer…

UVA 1394 And Then There Was One / Gym 101415A And Then There Was One / UVAlive 3882 And Then There Was One / POJ 3517 And Then There Was One / Aizu 1275 And Then There Was One (动态规划,思维题) Description Let's play a stone removing game. Initially, n ston…

题目 https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1736 题意 汉诺塔问题,给定初始状态和最终状态,求最小步数(一定可行) 思路 本来以为是搜索, 如刘书思路. 由于只有三根柱子,假设n-1到i + 1号盘子都已经对齐,对于没有对齐的i号盘子,必要有两根柱子用于转移,一根是现在i所在的柱子,另一根是目的柱子,第三根空闲柱子存放其他无用…

10067 - Playing with Wheels 题目页:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1008 从一开始思路就不对,之后才焕然大悟……每次都是这样. 还有,感觉搜索和图遍历有点分不清呢. 在第63行加入 if (u == target) return; 可以提速很多,可以从300ms左右降低到100ms以内. ?…