HDU_6043_KazaQ & #39; s Socks,

HDU_6043_KazaQ's 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

Problem Description KazaQWears socks everyday.

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 KazaQHas to wash them. These socks will be put in the closets again in tomorrow evening.

KazaQWocould like to know which pair of socks he shocould wear on the k-th day.

 

InputThe input consists of multiple test cases. (about 2000)

For each case, there is a line contains two numbers n, k (2 ≤ n ≤ limit, 1 ≤ k ≤ 1018 ).

 

OutputFor 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 Input3 73 64 9

 

Sample OutputCase #1: 3 Case #2: 1 Case #3: 2

 

Source2017 Multi-University Training Contest-Team 1

• Cycle, find the rule
• The first n orders remain unchanged, and the cycle starting from n + 1 is 2 N-2.
• Take n = 4 as an example.
• 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 }