You must have heard that the Chinese culture is quite different from that of Europe or Russia. So some Chinese habits seem quite unusual or even weird to us. So it is known that there is one popular game of Chinese girls. N girls stand forming a circ…

题目传送门 /* 杭电一题(ACM_steps 2.2.4)的升级版,使用到高精度: 这次不是简单的猜出来的了,求的是GCD (n, k) == 1 最大的k(1, n/2): 1. 若n是奇数,则k = (n-1) / 2: 2. 若n是偶数,讨论(n-1)/2 的奇偶性,若不是奇数,则是n/2-2: 详细解释(证明):http://www.xuebuyuan.com/1552889.html */ #include <cstdio> #include <iostream> #i…

A - Chinese Girls' Amusement Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description       You must have heard that the Chinese culture is quite different from that of Europe or Russia. So som…

 Chinese Girls' Amusement Time Limit:1000MS     Memory Limit:64000KB     64bit IO Format:%lld & %llu Submit Status Practice ACdream 1210 Description       You must have heard that the Chinese culture is quite different from that of Europe or Russia.…

Chinese Girls' Amusement Time Limit: 2000/1000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others) SubmitStatisticNext Problem Problem Description You must have heard that the Chinese culture is quite different from that of Europe or Russia. So…

Description       You must have heard that the Chinese culture is quite different from that of Europe or Russia. So some Chinese habits seem quite unusual or even weird to us.       So it is known that there is one popular game of Chinese girls. N gi…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1313 题目意思:有 N 个人(编号依次为1~N)围成一个圆圈,要求求出最大的 K (1 ≤ K ≤ N/2),表示从编号为1的人开始,将球传递给他后一个人数起的第K个人,第K个人又传递给往后数的第K个人......要求这样传递下去,且每个人都有机会接到球.也就是不存在当未使得全部人都接到一次球的情况下,某个人接收到两次以上的球. 详细的解题报告在这里: http:/…

题意:有n个女孩围成一个圈从第1号女孩开始有一个球,可以往编号大的抛去(像传绣球一样绕着环来传),每次必须抛给左边第k个人,比如1号会抛给1+k号女孩.给出女孩的人数,如果他们都每个人都想要碰到球一次,那么这个k应该是多少(满足 1 ≤ K ≤ N/2 且 k必须尽量大)?   例如:n=7,那么1号开始拿球,抛球的顺序是 1, 4, 7, 3, 6, 2, 5, 1.  当球重新回到1女孩手中时,每个人刚好只玩了一次.注:这个数字相当大(3 ≤ N ≤ 102000) 思路: 方法(1): 暴…

/* 实际上就是求一个k,满足k<=n/2,且gcd(n,k)=1 如果n为奇数,k为[n/2] 如果n为偶数,k=n/2-1-(n/2)%2 */ #include <iostream> using namespace std; string s; void div2() { string t; int l = s.size() - 1, tem = s[0] - '0'; if (tem > 1) t += '0' + tem / 2; tem &= 1; for (i…

解题报告:n个人围坐成一圈,并且将这n个人从1到n编号,然后编号为1 的人手上有一个物品,将这个物品往向左传递给第k个人,1<=k<=n/2,当这个物品再次传到编号为1 的人的手上时,游戏结束,要使这个k最大,并且,在游戏结束后要求每一个人都要拿过这个物品,求这个最大的k. 看了一下就猜想大概就是在n/2附近去找,然后做了几个小的数字分析了一下,发现猜想没有错误,不过不能证明,这题还有一个到现在还是很困惑的地方,为什么我的代码数组一定要开到20000,本来开到2000就够了,但我的一定要开到2…