Joseph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2126 Accepted Submission(s): 1291
Problem Description
The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, standing in circle every mth is going to be executed and only the life of the last remaining person will be saved. Joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. For example when n = 6 and m = 5 then the people will be executed in the order 5, 4, 6, 2, 3 and 1 will be saved.
Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy.
Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy.
Input
The input file consists of separate lines containing k. The last line in the input file contains 0. You can suppose that 0 < k < 14.
Output
The output file will consist of separate lines containing m corresponding to k in the input file.
Sample Input
34
0
Sample Output
530
思路：一开始看到数据很小，打表写的用vector容器模拟，这样模拟肯定超时，但由于只到13，所以，把数据打出来直接上表。
另一种方法是只记录好人的开头和结尾，然后每次出一个人就更新开头和结尾，如果出来的人在头和尾之间就不可行。
关键是更新开头和结尾： head=((((head-(p+1)%i)+1+i))%i);
wei=((((wei-(p+1)%i)+i+1))%i);（p+1）%i为去掉前面的数，然后开始的第一个数的原下标，然后算出head,与原下标之间的距离
(head-(p+1)%i+i)%i，因为新的开头为1，所以加1就为当前标号head=((((head-(p+1)%i)+1+i))%i);
1 #include<stdio.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<string.h>
5 #include<math.h>
6 #include<vector>
7 void run();
8 using namespace std;
9 vector<int>my; int yy[14];
10 int main(void)
11 {
12 int n,i,j,k,p,q;run();
13 while(scanf("%d",&p),p!=0)
14 {
15 printf("%d ",yy[p]);
16 }
17 return 0;
18 }
19 void run()
20 {
21
22 int n,i,j,k,p,q;
23 for(k=1;k<14;k++)
24 {
25 for(j=k+1;; j++)
26 {
27 int head=1;
28 int wei=k;
29 int sum=0;
30 for(i=2*k; i>=1; i--)
31 { p=(j)%i;
32 if(p==0)
33 p=i;
34 if(p>=head&&p<=wei)
35 break;
36 sum++;
37 head=((((head-(p+1)%i)+1+i))%i);
38 wei=((((wei-(p+1)%i)+i+1))%i);
39 if(head==0)head=i-1;
40 if(wei==0)wei=i-1;
41 }
42 if(sum==k)
43 break;
44 }
45 yy[k]=j;
46 }
47 }