KazaQ's Socks Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1890Accepted Submission(s): 1061 Problem Description KazaQ wears socks everyday. At the beginning, he has n pairs of socks
KazaQ's Socks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1890 Accepted Submission(s): 1061
At the beginning, he has n pairs of socks numbered from 1 to n in his closets.
Every morning, he puts on a pair of socks which has the smallest number in the closets.
Every evening, he puts this pair of socks in the basket. If there are n−1 pairs of socks in the basket now, lazy KazaQ has to wash them. These socks will be put in the closets again in tomorrow evening.
KazaQ would like to know which pair of socks he should wear on the k-th day.
Input The input consists of multiple test cases. (about 2000)
For each case, there is a line contains two numbers n,k (2≤n≤109,1≤k≤1018).
Output For each test case, output "Case #x: y" in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.
Sample Input 3 7 3 6 4 9
Sample Output Case #1: 3 Case #2: 1 Case #3: 2
Source 2017 Multi-University Training Contest - Team 1
- 循环，找规律
- 前n个顺序不变，从第n+1个开始长度为2n-2的循环
- 拿n=4为例
- 1 2 3 4 /1 2 3 /1 2 4 /1 2 3 /1 2 4 ....
1 #include <iostream> 2 #include <string> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 #include <climits> 7 #include <cmath> 8 #include <vector> 9 #include <queue> 10 #include <stack> 11 #include <set> 12 #include <map> 13 using namespace std; 14 typedef long long LL ; 15 typedef unsigned long long ULL ; 16 const int maxn = 1e5 + 10 ; 17 const int inf = 0x3f3f3f3f ; 18 const int npos = -1 ; 19 const int mod = 1e9 + 7 ; 20 const int mxx = 100 + 5 ; 21 const double eps = 1e-6 ; 22 const double PI = acos(-1.0) ; 23 24 int main(){ 25 // freopen("in.txt","r",stdin); 26 // freopen("out.txt","w",stdout); 27 LL T=0, n, k, ans; 28 while(~scanf("%lld %lld",&n,&k)){ 29 if(k<=n){ 30 ans=k; 31 }else{ 32 k-=n; 33 LL m=k/(n-1); 34 if(m*(n-1)<k)m++; 35 k=k-(n-1)*(m-1); 36 ans=k; 37 if(!(m&1)) 38 if(k==n-1) 39 ans++; 40 41 } 42 printf("Case #%lld: %lld\n",++T,ans); 43 } 44 return 0; 45 }