HDU-5668-Circle（中国余数定理/解同余方程组）

Circle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 320    Accepted Submission(s): 102

Problem Description

     Satiya August is in charge of souls.

     He finds n souls,and lets them become a circle.He ordered them to play Joseph Games.The souls will count off from the soul 1 .The soul who is numbered k will be taken out,and will not join in the game again.

     Now Satiya August has got the sequence in the Out Ordered,and ask you the smallest k .If you cannot give him a correct answer,he will kill you!

 

Input

     The first line has a number T,means testcase number.

     Each test,first line has a number n .

     The second line has n numbers,which are the sequence in the Out Ordered**(The person who is out at aith round was numbered i )**.

     The sequence input must be a permutation from 1 to n .

    1≤T≤10,2≤n≤20 .

 

Output

     For each case,If there is a eligible number k ,output the smallest k ,otherwise,output”Creation August is a SB!”.

 

Sample Input

   
   
   
   

    
    
    
    1
7
7 6 5 4 3 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    420
   
   
   
   

 

Source

BestCoder Round #80

 

考虑对给定的出圈序列进行一次模拟,对于出圈的人我们显然可以由位置,编号等关系得到一个同余方程 一圈做下来我们就得到了n个同余方程 对每个方程用扩展欧几里得求解,最后找到最小可行解就是答案. 当然不要忘了判无解的情况.

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <stack>
using namespace std;
const int MAXN = 107;
int num[MAXN];
long long mod[MAXN],b[MAXN];  // k(%mod)==b
bool vis[MAXN];
void egcd(long long a, long long b, long long &d, long long &x, long long &y)
{
    if (!b)d=a,x=1,y=0;
    else
    {
        egcd(b, a%b, d, y, x);
        y -= x*(a / b);
    }
}
long long CRT(int n)
{
    long long M=mod[1],B=b[1],x,y,d;
    for (int i=2; i<=n; i++)
    {
        egcd(M,mod[i],d,x,y);
        if ((b[i]-B)%d)return -1;
        x=(b[i]-B)/d*x%(mod[i]/d);
        B+=x*M;
        M=M/d*mod[i];
        B%=M;
    }
    return (B+M)%M?(B+M)%M:M;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n,p;
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        for(int i=1; i<=n; ++i)
        {
            scanf("%d",&p);
            num[p]=i;
        }
        int j=1,k;
        for(int i=1; i<=n; ++i)
        {
            k=1;
            while(j!=num[i])
            {
                if(!vis[j])++k;
                j=j%n+1;
            }
            vis[num[i]]=1;
            mod[i]=n-i+1;
            b[i]=k%mod[i];
        }
        int flag=CRT(n);
        if(flag==-1)cout<<"Creation August is a SB!\n";
        else cout<<flag<<endl;
    }
    return 0;
}