Joseph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2231 Accepted Submission(s): 1354
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
3
4
0
Sample Output
5
30
Source
Recommend
#include <cstdio> int kill(int n, int k, int i) { if(i==1) return (n+k-1)%n; else return (kill(n-1, k, i-1)+k)%n; } int f[15]; void dabiao() { int i, m, k; for(k=1; k<14; k++) { for(m=k; ;++m) { for(i=1; i<=k; i++) { if(kill(2*k, m, i)+1<=k) break; } if(i>k) break; } f[k]=m; } } int main() { dabiao(); int t, k, n, i, j; while(scanf("%d", &k), k) { printf("%d ", f[k]); } return 0; }