hdu 5668 Circle 中国剩余定理

Circle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

 思路：先模拟将出队的顺序求出，再将同余方程求出，不互质的中国剩余定理；

　　　例：7 

　　　　　6 7 5 3 1 2  4

             出队顺序：5 6 4 7 3 1 2 ；

　　　 　 暴力约瑟夫得到；

　　　　   5≡k mod (n);

　　　　　1≡k mod (n-1);

              5≡k mod (n-2);

　　　　　1≡k mod (n-3);

              3≡k mod (n-4);

　　　　　1≡k mod (n-5);

　　　　   1≡k mod (n-6);

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define mod 1000000007
#define inf 999999999
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF ) return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
ll xu[100010];
ll pos[100010];
ll a[100010];
ll b[100010];
ll flag[110];
ll gcd(ll x,ll y)
{
    if(x%y==0)
    return y;
    else
    return gcd(y,x%y);
}
void exgcd(ll a, ll b, ll &x, ll &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return;
    }
    exgcd(b, a % b, x, y);
    ll tmp = x;
    x = y;
    y = tmp - (a / b) * y;
}
int main()
{
    ll x,y,z,i,t;
    scanf("%lld",&x);
    while(x--)
    {
        memset(a,0,sizeof(a));
        memset(flag,0,sizeof(flag));
        scanf("%lld",&y);
        for(i=0;i<y;i++)
        {
            scanf("%lld",&xu[i]);
            pos[xu[i]]=i+1;
        }
        int num=y;
        for(i=1;i<=y;i++)
        b[i]=num--;
        int st=1;
        for(i=1;i<=y;i++)
        {
            while(1)
            {
                if(st==pos[i])
                break;
                if(!flag[st])a[i]++;
                st++;
                if(st==y+1)
                st=1;
            }
            a[i]++;
            flag[st]=1;
            //cout<<i<<" "<<a[i]<<endl;
        }
        ll a1=a[1],b1=b[1];
        ll jie=1;
        for(i=2;i<=y;i++)
        {
            ll a2=a[i],b2=b[i];
            ll xx,yy;
            ll gys=gcd(b1,b2);
            if((a2-a1)%gys)
            {
                jie=0;
                break;
            }
            exgcd(b1,b2,xx,yy);
            xx=(xx*(a2-a1))/gys;
            ll gbs=b1*b2/gys;
            a1=(((xx*b1+a1)%gbs)+gbs)%gbs;
            b1=gbs;
        }
        if(!jie)
        printf("Creation August is a SB!
");
        else if(a1!=0)
        printf("%lld
",a1);
        else
        printf("%lld
",b1);
    }
    return 0;
}