poj3708(公式化简+大数进制装换+线性同余方程组)

刚看到这个题目，有点被吓到，毕竟自己这么弱。

分析了很久，然后发现m，k都可以唯一的用d进制表示。也就是用一个ai，和很多个bi唯一构成。

这点就是解题的关键了。 之后可以发现每次调用函数f(x),相当于a(ai),b(bi)了一下。这样根据置换的一定知识，一定会出现循环，而把循环的大小看成取模，把从m->k的看成余，于是可以建立一个线性同余方程。

直接用模板解决之。。

 

Recurrent Function

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1102		Accepted: 294

Description

Dr. Yao is involved in a secret research on the topic of the properties of recurrent function. Some of the functions in this research are in the following pattern:

in which set {a_i} = {1, 2, …, d-1} and {b_i} = {0, 1, …, d-1}.
We denote:

Yao's question is that, given two positive integer m and k, could you find a minimal non-negative integer x that

Input

There are several test cases. The first line of each test case contains an integer d (2≤d≤100). The second line contains 2d-1 integers: a₁, …, a_d-1, followed by b₀, ..., b_d-1. The third line contains integer m (0<m≤10¹⁰⁰), and the forth line contains integer k (0<k≤10¹⁰⁰). The input file ends with integer -1. 

Output

For each test case if it exists such an integer x, output the minimal one. We guarantee the answer is less than 2⁶³. Otherwise output a word "NO". 

Sample Input

2
1
1 0
4
7
2
1
0 1
100
200
-1

Sample Output

1
NO

Hint

For the first sample case, we can see that f(4)=7. And for the second one, the function is f(i)=i. 

//
//  main.cpp
//  poj3708
//
//  Created by 陈加寿 on 15/11/28.
//  Copyright (c) 2015年 陈加寿. All rights reserved.
//

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;


int a[110],b[110];
char strm[110],strk[110];
int m[110],k[110];
int savem[1100],savek[1100];
int cntm,cntk;

void Tentod(int ten[],int len,int &cnt,int d,int save[])
{
    int tcnt=0;
    while(ten[tcnt]==0) tcnt++;
    cnt=0;
    while(tcnt<len)
    {
        for(int i=tcnt;i<len;i++)
        {
            ten[i+1] += (ten[i]%d)*10;
            ten[i] /= d;
        }
        save[cnt++] = ten[len]/10;
        ten[len]=0;
        while(tcnt<len&&ten[tcnt]==0) tcnt++;
    }
    /*
    for(int i=0;i<cnt;i++)
        printf("%d ",save[i]);
    printf("\n");
    */
}



/*对于x=r0(mod m0)
 x=r1(mod m1)
 ...
 x=rn(mod mn)
 输入数组m和数组r，返回[0,[m0,m1,...,mn]-1] 范围内满足以上等式的x0。
 x的所有解为:x0+z*[m0,m1,...mn](z为整数)
 */
long long cal_axb(long long a,long long b,long long mod)
{
    //防乘法溢出
    long long sum=0;
    while(b)
    {
        if(b&1) sum=(sum+a)%mod;
        b>>=1;
        a=(a+a)%mod;
    }
    return sum;
}

//ax + by = gcd(a,b)
//传入固定值a,b.放回 d=gcd(a,b), x , y
void extendgcd(long long a,long long b,long long &d,long long &x,long long &y)
{
    if(b==0){d=a;x=1;y=0;return;}
    extendgcd(b,a%b,d,y,x);
    y -= x*(a/b);
}

long long Multi_ModX(long long m[],long long r[],int n)
{
    long long m0,r0;
    m0=m[0]; r0=r[0];
    for(int i=1;i<n;i++)
    {
        long long m1=m[i],r1=r[i];
        long long k0,k1;
        long long tmpd;
        extendgcd(m0,m1,tmpd,k0,k1);
        if( (r1 - r0)%tmpd!=0 ) return -1;
        k0 *= (r1-r0)/tmpd;
        m1 *= m0/tmpd;
        r0 = ( cal_axb(k0,m0,m1)+r0)%m1;
        m0=m1;
    }
    return (r0%m0+m0)%m0;
}

int main(int argc, const char * argv[]) {
    int d;
    while(cin>>d)
    {
        if(d==-1) break;
        for(int i=1;i<d;i++) cin>>a[i];
        for(int i=0;i<d;i++) cin>>b[i];
        scanf("%s",strm);
        scanf("%s",strk);
        int len = strlen(strm);
        for(int i=0;i<len;i++)
            m[i] = strm[i]-'0';
        Tentod(m,len,cntm,d,savem);
        len = strlen(strk);
        for(int i=0;i<len;i++)
            k[i] = strk[i]-'0';
        Tentod(k, len, cntk, d, savek);
        // 这样就得到了。a，b。。。
        // 然后构建同模方程
        if(cntm != cntk)
        {
            printf("NO\n");
        }
        else
        {
            int flag=0;
            long long m[1010],r[1010];
            for(int i=0;i<cntm-1;i++)
            {
                int a1=savem[i],a2=savek[i];
                int tm=1;
                int ta=a1;
                while(b[ta]!=a1)
                {
                    tm++;
                    ta=b[ta];
                }
                ta=a1;
                int tr=0;
                while(ta != a2)
                {
                    tr++;
                    ta=b[ta];
                    if(ta==a1)
                    {
                        flag=1;
                        break;
                    }
                }
                m[i]=tm;
                r[i]=tr;
                if(flag==1) break;
            }//这里面都是b
            int a1=savem[cntm-1],a2=savek[cntm-1];
            int tm=1;
            int ta=a1;
            while(a[ta]!=a1)
            {
                tm++;
                ta=a[ta];
            }
            ta=a1;
            int tr=0;
            while(ta != a2)
            {
                tr++;
                ta=a[ta];
                if(ta==a1)
                {
                    flag=1;
                    break;
                }
            }
            m[cntm-1]=tm;
            r[cntm-1]=tr;
            if(flag == 1)
            {
                printf("NO\n");
            }
            else
            {
                long long ans=Multi_ModX(m, r,cntm);
                if(ans==-1) printf("NO\n");
                else cout<<ans<<endl;
            }
        }
    }
    return 0;
}