UVA10940 - Throwing cards away II(找规律) 题目链接 题目大意:桌上有n张牌,依照1-n的顺序从上到下,每次进行将第一张牌丢掉,然后把第二张放到这叠牌的最后.重复进行这种操作.直到仅仅剩下一张牌. 解题思路:仅仅能先暴力.将前面小点的n打印出来.看看有什么规律. 规律:f[2^k + mod] = 2*mod;(mod > 0); n = 1须要特判. 代码: #include <cstdio> #include <cstring> cons…

题意略: 先暴力打表发现规律 N=1 ans=1N=2 ans=2N=3 ans=2N=4 ans=4N=5 ans=2N=6 ans=4N=7 ans=6N=8 ans=8N=9 ans=2N=10 ans=4N=11 ans=6N=12 ans=8N=13 ans=10N=14 ans=12N=15 ans=14N=16 ans=16N=17 ans=2N=18 ans=4N=19 ans=6N=20 ans=8N=21 ans=10N=22 ans=12N=23 ans=14N=24 an…

Throwing cards away I   Given is an ordered deck of n cards numbered 1 to n with card 1 at the top and card n at the bottom. The following operation is performed as long as there are at least two cards in the deck:  Throw away the top card and move t…

原题 Throwing cards away I   Given is an ordered deck of n cards numbered 1 to n with card 1 at the top and card n at the bottom. The following operation is performed as long as there are at least two cards in the deck:   Throw away the top card and mo…

题目: Problem B: Throwing cards away I Given is an ordered deck of n cards numbered 1 to n with card 1 at the top and card n at the bottom. The following operation is performed as long as there are at least two cards in the deck: Throw away the top car…

书上具体所有题目:http://pan.baidu.com/s/1hssH0KO 代码:(Accepted,0 ms) //UVa10935 - Throwing cards away I #include<iostream> #include<queue> int N; int main() { //freopen("in.txt", "r", stdin); while (scanf("%d", &N) !=…

 Throwing cards away I Given is an ordered deck of  n  cards numbered 1 to n  with card 1 at the top and card  n  at the bottom. The following operation is performed as long as there are at least two cards in the deck: Throw away the top card and m…

D - Throwing cards away I Given is an ordered deck of n cards numbered 1 to n with card 1 at the top and card n at the bottom. The following operation is performed as long as there are at least two cards in the deck:   Throw away the top card and mov…

/** 题目:G - Harmonic Number (II) 链接:https://vjudge.net/contest/154246#problem/G 题意:给定一个数n,求n除以1~n这n个数的和.n达到2^31 - 1; 思路: 首先我们观察一下数据范围,2^31次方有点大,暴力会超时,所以我们看看有没有啥规律,假设 tmp 是 n/i 的值,当n == 10的时候(取具体值) 当 tmp = 1 时,个数 是10/1 - 10/2 == 5个 当 tmp = 2 时,个数 是10/2…

题意:给定一个表达式,然后让你求表达式的值. 析:多写几个就会发现规律. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring>…