2017 Multi-University Training Contest - Team 1 簽到題 ...
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
- 迴圈，找規律
- 前n個順序不變，從第n+1個開始長度為2n-2的迴圈
- 拿n=4為例
- 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 }
