2017 Multi-University Training Contest

KazaQ's Socks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 338    Accepted Submission(s): 212


Sample Input
3 7
3 6
4 9
Sample Output
Case #1: 3
Case #2: 1
Case #3: 2
Source
分析:
然后分别去判断即可!详细请看代码。
下面给出AC代码:
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 int main()
 5 {
 6     ll n,m;
 7     ll p=1;
 8     while(scanf("%lld%lld",&n,&m)!=EOF)
 9     {
10         printf("Case #%lld: ",p++);
11         if(m<=n)
12         {
13             printf("%lld
",m);
14             continue;
15         }
16         ll c=m-n;
17         if(c%(2*(n-1))==0)
18         {
19             printf("%lld
",n);
20             continue;
21         }
22         if(c%(2*(n-1))<=n-1)
23             printf("%lld
",c%(2*(n-1)));
24         else
25         {
26             ll k=c%(2*(n-1))%(n-1);
27             printf("%lld
",k);
28         }
29     }
30     return 0;
31 }
原文地址:https://www.cnblogs.com/ECJTUACM-873284962/p/7237861.html