Joseph's question

Problem description: n people (No. 0 ~ (N-1), reporting starts from 0, reporting ends from m-1), and reporting continues from 0. Calculate the number of the winner.

After the first person (number must be (S-1) mod n) is listed, the remaining n-1 individuals form a new Joseph ring (starting with the person numbered k = m mod n ):
K + 1 k + 2... N-2, n-1, 2,... K-2
And 0 is reported from k.
Now let's convert their numbers:
K --> 0
K + 1 --> 1
K + 2 --> 2
...
...
K-2> N-2
After the transformation, it will completely become a subproblem of (n-1) the number of individual reports. If we know the solution of this subproblem: for example, x is the final winner, so it's just n people's situation to change x back Based on the table above ?!! The formula for changing back is very simple. I believe everyone can introduce it: x' = (x + k) mod n
How can I solve the problem of (n-1) number of personal reports? Yes, as long as you know (n-2) the individual's solution. (N-2) What about personal solutions? Of course it is the first (n-3) situation ---- this is obviously a reverse push problem! Well, the idea is coming out. The recursive formula is as follows:
F indicates that I personally play the game and report m to exit the final winner number. The final result is f [n].
Recurrence Formula
F [1] = 0;
F = (f + m) mod I; (I> 1)
With this formula, we need to calculate the value of f from the 1-n order, and the final result is f [n]. Because the number in real life always starts from 1, we Output f [n] + 1

Sub-questions about Joseph.

1. Poj 3517 And Then There Was One

Fixed start point is not 1, but m.

First remove a number, convert it to the n-1 Number of the Joseph Ring problem, and then convert the final result s = (m + s) % n + 1.

 

 

#include <stdio.h>  #include <stdlib.h>  #include <algorithm>  #include <string.h>  #include <math.h>  #include <iostream>  using namespace std;  #define Maxn 10100   int f[Maxn]; int main() {     #ifndef ONLINE_JUDGE      freopen("in.txt","r",stdin);     #endif      int n,m,k;     while(scanf(" %d %d %d",&n,&k,&m)!=EOF)     {         if(n == 0 && m == 0 && k == 0) break;         f[1] = 0;         for(int i=2;i<=n;i++)         {             f[i] = (f[i-1] + k)%i;         }         printf("%d\n",(f[n-1]+m)%n + 1);     }     return 0; } #include <stdio.h>#include <stdlib.h>#include <algorithm>#include <string.h>#include <math.h>#include <iostream>using namespace std;#define Maxn 10100int f[Maxn];int main(){    #ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    #endif    int n,m,k;    while(scanf(" %d %d %d",&n,&k,&m)!=EOF)    {        if(n == 0 && m == 0 && k == 0) break;        f[1] = 0;        for(int i=2;i<=n;i++)        {            f[i] = (f[i-1] + k)%i;        }        printf("%d\n",(f[n-1]+m)%n + 1);    }    return 0;}

2. Hoj 1016 Joseph's problem I

Each interval is a prime number. We only need to pre-screen the prime number within a range.

 

?#include <stdio.h>  #include <stdlib.h>  #include <algorithm>  #include <string.h>  #include <math.h>  #include <iostream>  using namespace std;  #define Maxn 50000   int prime[Maxn]; int vis[Maxn]; int get_Prime(int n) {     memset(vis,0,sizeof(vis));     int np = 0;     for(int i=2;i<=n;i++)     {         if(!vis[i]) prime[np++] = i;         long long t;         for(int j=0;j<np && (t = prime[j]*i)<=n;j++)         {             vis[t] = 1;             if(i%prime[j] == 0) break;         }     } }  int f[Maxn]; int main() {     #ifndef ONLINE_JUDGE      freopen("in.txt","r",stdin);     #endif      get_Prime(Maxn);     int n;     while(scanf(" %d",&n)!=EOF && n!=0)     {         f[1] = 0;         for(int i=n-2;i>=0;i--)         {             f[n-i] = (f[n-i-1] + prime[i]) % (n - i);         }         printf("%d\n",f[n]+1);     }     return 0; } #include <stdio.h>#include <stdlib.h>#include <algorithm>#include <string.h>#include <math.h>#include <iostream>using namespace std;#define Maxn 50000int prime[Maxn];int vis[Maxn];int get_Prime(int n){    memset(vis,0,sizeof(vis));    int np = 0;    for(int i=2;i<=n;i++)    {        if(!vis[i]) prime[np++] = i;        long long t;        for(int j=0;j<np && (t = prime[j]*i)<=n;j++)        {            vis[t] = 1;            if(i%prime[j] == 0) break;        }    }}int f[Maxn];int main(){    #ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    #endif    get_Prime(Maxn);    int n;    while(scanf(" %d",&n)!=EOF && n!=0)    {        f[1] = 0;        for(int i=n-2;i>=0;i--)        {            f[n-i] = (f[n-i-1] + prime[i]) % (n - i);        }        printf("%d\n",f[n]+1);    }    return 0;}3.Hoj 1107 Joseph's problem II

K good guys and k bad guys cannot be killed each time. Ask the smallest m.

Solution: similar to array simulation. The range of start and end for each change. Each time a person is killed. The next number is 0.

include <stdio.h>  #include <stdlib.h>  #include <algorithm>  #include <string.h>  #include <math.h>  #include <iostream>  using namespace std;  #define Maxn 10100   int f[15]; bool solve(int k,int m) {     int start = 0,end = k - 1;     bool flag = true;     for(int i=2*k;i>k;i--)     {         int kill = (m-1)%i;         if(kill>=start && kill<=end)         {             flag = false;             break;         }         start = ((start - m)%i + i)%i;         end = ((end - m)%i + i)%i;     }     return flag; } void init() {     for(int k=1;k<14;k++)     {         for(int m=k+1;;m++)         {             if(solve(k,m))             {                 f[k] = m;                 break;             }         }     } } int main() {     #ifndef ONLINE_JUDGE      freopen("in.txt","r",stdin);     #endif      init();     int k;     while(scanf(" %d",&k)!=EOF && k!=0)     {         printf("%d\n",f[k]);     }     return 0; } #include <stdio.h>#include <stdlib.h>#include <algorithm>#include <string.h>#include <math.h>#include <iostream>using namespace std;#define Maxn 10100int f[15];bool solve(int k,int m){    int start = 0,end = k - 1;    bool flag = true;    for(int i=2*k;i>k;i--)    {        int kill = (m-1)%i;        if(kill>=start && kill<=end)        {            flag = false;            break;        }        start = ((start - m)%i + i)%i;        end = ((end - m)%i + i)%i;    }    return flag;}void init(){    for(int k=1;k<14;k++)    {        for(int m=k+1;;m++)        {            if(solve(k,m))            {                f[k] = m;                break;            }        }    }}int main(){    #ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    #endif    init();    int k;    while(scanf(" %d",&k)!=EOF && k!=0)    {        printf("%d\n",f[k]);    }    return 0;}