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
2017 Multi-University Training Contest - Team 1
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6043
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\n",m);
14 continue;
15 }
16 ll c=m-n;
17 if(c%(2*(n-1))==0)
18 {
19 printf("%lld\n",n);
20 continue;
21 }
22 if(c%(2*(n-1))<=n-1)
23 printf("%lld\n",c%(2*(n-1)));
24 else
25 {
26 ll k=c%(2*(n-1))%(n-1);
27 printf("%lld\n",k);
28 }
29 }
30 return 0;
31 }