思路: 找每一个数的循环节,注意优化!! 每次找一个数的循环节时,记录其路径,下次对应的数就不用再找了…… 代码如下: #include<iostream> #include<cstdio> #include<stack> #include<cstring> #define ll __int64 using namespace std; ],to[],an[]; stack<int>p; ll gcd(ll a,ll b) { if(a<b…

首先将扑克牌进行一次置换,然后分解出所有的循环节,所有循环节的扑克牌个数的最小公倍数即为答案 #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; #define LL long long int n,k; LL d[850],ct; int vis[850]; int a[850]; int dfs(int x) { if(!vis[a[x]]) { ct++;…

Double Dealing Time Limit: 50000/20000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1924    Accepted Submission(s): 679 Problem Description Take a deck of n unique cards. Deal the entire deck out to k players in…

找每一位的循环节.求lcm Double Dealing Time Limit: 50000/20000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1806    Accepted Submission(s): 622 Problem Description Take a deck of n unique cards. Deal the entire deck out to…

# include <stdio.h> # include <algorithm> # include <string.h> using namespace std; __int64 gcd(__int64 a,__int64 b) { if(b==0) return a; return gcd(b,a%b); } int main() { int n,k,i,j,vis[810],m,num[810],x; __int64 res,cot; while(~scanf(…

题目:有N个正实数(注意是实数,大小升序排列) x1 , x2 ... xN,另有一个实数M. 需要选出若干个x,使这几个x的和与 M 最接近. 请描述实现算法,并指出算法复杂度. 代码如下: #include<iostream> using namespace std; int min_diff(int data[],int n,int &min_i,int &min_j,int number); int main() { int number,n,i; cin>>…

Least Common Multiple Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 53016    Accepted Submission(s): 20171 Problem Description The least common multiple (LCM) of a set of positive integers is…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1908 Double Queue Description The new founded Balkan Investment Group Bank (BIG-Bank) opened a new office in Bucharest, equipped with a modern computing environment provided by IBM Romania, and using mod…

题目链接: https://cn.vjudge.net/problem/POJ-2429 题目大意: 给出两个数的gcd和lcm,求原来的这两个数(限定两数之和最小). 解题思路: 首先,知道gcd和lcm求原来的两个数,需要分解lcm / gcd .将其分解为互质的两个数. 首先将lcm/gcd质因数分解,要分解出沪互质两个数字,那么这两个数字的gcd=1,也就是没有公共的质因子,所以可以直接枚举这两个数字的质因子,如果一个数要取这个质因子,就把它的指数全部取掉. 质因数分解用大数因式分解来做…

Fibonacci Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4844    Accepted Submission(s): 2245 Problem Description 2007年到来了.经过2006年一年的修炼,数学神童zouyu终于把0到100000000的Fibonacci数列(f[0]=0,f[1]=1;f[i] =…