Circle

    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 soul1.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!

    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.

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

1
7
7 6 5 4 3 2 1

420

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;
}

HDU-5668-Circle（中国余数定理/解同余方程组）

原文：http://blog.csdn.net/qq978874169/article/details/51253717