这道题是有规律的博弈题目,,, 所以我们只需要找出规律来就ok了 牛人用sg函数暴力找规律,菜鸟手工模拟以求规律...[牢骚] if(m>=2) { if(n<=m) {first第一口就可以吃掉所有的.所以first必赢,} else {first无法一口吃掉所有的,所以second成了主动的了,如果first第一口吃掉k1个,那么明智的second只要吃掉k2个就可以了(n-k1-k2是偶数,也包括 0的),使得 剩下的数字是分成两个数字数目相等的堆,以后的工作便是first做什么,那么s

已知有n个单位的水,问有几种方式把这些水喝完,每天至少喝1个单位的水,而且每天喝的水的单位为整数.看上去挺复杂要跑循环,但其实上,列举几种情况之后就会发现是找规律的题了= =都是2的n-1次方,而且这题输出二进制数就行了......那就更简单了,直接输出1,然后后面跟n-1个0就行了╮(╯_╰)╭ 下面AC代码 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm>

2014多校 第八题 1008 2014 Multi-University Training Contest 8 4952 Number Transformation Number Transformation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 85 Accepted Submission(s): 31 Problem Descr

http://codeforces.com/contest/456/problem/B CF#260 div2 B Fedya and Maths Codeforces Round #260 B. Fedya and Maths time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Fedya studies in a gymnasi

题意:找字串中最长回文串的最小值的串 m=2的时候暴力打表找规律,打表可以用二进制枚举

题目传送门 /* 水题 找规律输出 */ #include <cstdio> #include <iostream> #include <cstring> #include <string> #include <map> #include <algorithm> #include <vector> #include <set> #include <cmath> using namespace std

Javabeans are delicious. Javaman likes to eat javabeans very much. Javaman has n boxes of javabeans. There are exactly i javabeans in the i-th box (i = 1, 2, 3,...n). Everyday Javaman chooses an integer x. He also chooses several boxes where the numb

递归思想的条件:1.函数自己调用自己 2.函数必须有一个固定的返回值(如果没有这个条件会发生死循环) ----规律很重要 简单递归题目一: 设计一个函数计算一个整数的n次方,比如2的3次方,就是8 步骤: 1.定义方法,以及方法中的参数--power( int base ,int n ) 2.根据定义的方法来找规律: 函数终止条件:n==0时,返回1:n<0时,返回-1:红色箭头表示函数的执行过程,蓝色箭头表示函数达到条件后开始一层一层的返回值,具体代码如下: long power(int ba

讨厌博弈,找规律找半天还是错的.... 1228: [SDOI2009]E&D Time Limit: 10 Sec Memory Limit: 162 MB Submit: 666 Solved: 380 [Submit][Status][Discuss] Description 小E 与小W 进行一项名为"E&D"游戏.游戏的规则如下:桌子上有2n 堆石子,编号为1..2n.其中,为了方便起见,我们将第2k-1 堆与第2k 堆(1 ≤ k ≤ n)视为同一组.第i堆

Nim or not Nim? Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 3032 Description Nim is a two-player mathematic game of strategy in which players take turns removing objects from distinct heaps.

题目传送门 /* 找规律/数位DP:我做的时候差一点做出来了,只是不知道最后的 is_one () http://www.cnblogs.com/crazyapple/p/3315436.html 数位DP:http://blog.csdn.net/libin56842/article/details/11580497 */ #include <cstdio> #include <iostream> #include <algorithm> #include <c

题目地址:http://acm.sgu.ru/problem.php?contest=0&problem=126 /* 找规律,智商不够,看了题解 详细解释:http://blog.csdn.net/ahfywff/article/details/7432524 */ #include <cstdio> #include <cmath> #include <cstring> #include <string> #include <iostrea

Problem about GCD Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 470    Accepted Submission(s): 77 Problem Description Given integer m. Find multiplication of all 1<=a<=m such gcd(a, m)=1 (cop

Description Farmer John's N cows (1 <= N <= 100,000) are lined up in a row.Each cow is labeled with a number in the range 1...K (1 <= K <=10,000) identifying her breed. For example, a line of 14 cows might have these breeds: 1 5 3 2 5 1 3 4 4

Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6012   Accepted: 3341 Description N children standing in circle who are numbered 1 through N clockwise are waiting their candies. Their teacher distributes the candies by in the following w

题目链接: http://codeforces.com/problemset/problem/353/D?mobile=true H. Queue time limit per test 1 secondmemory limit per test 256 megabytes 问题描述 There are n schoolchildren, boys and girls, lined up in the school canteen in front of the bun stall. The b

Number Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 152357    Accepted Submission(s): 37087 Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A

题目 //找规律,123321123321123321…发现这样排列恰好可以错开 // 其中注意题中数据范围: M是行,N是列,3 <= N < 2×M //则猜测:m,m,m-1,m-1,m-2,m-2,……,2,2,1,1求出前m个数字的和就是答案. //发现案例符合(之前的代码第二天发现案例都跑不对,真不知道我当时眼睛怎么了) #include <iostream> #include<stdio.h> #include<string.h> #inclu

题目 大意: 求二进制的a加到b的进位数. 思路: 列出前几个2进制,找规律模拟. #include <stdio.h> #include <iostream> #include <algorithm> #include <string.h> #include <math.h> #include <stack> #include <vector> using namespace std; int main() { int

题目链接:http://lightoj.com/volume_showproblem.php?problem=1245 题意就是求 n/i (1<=i<=n) 的取整的和这就是到找规律的题, i     1  2   3   4   5   6   7    8 a    8  4   2   2   1   1   1    1 你可以多写几组你会发现 有8-4个1:4-2个2:...其他例子也是这样: 当n = 10时 n/1 = 10, n/2 = 5说明(5, 10]这个前开后闭的区间