数学推导+高精度进制转换+解同余方程 POJ 3708

题意：点击打开链接

把m,k表示成d进制，对于这个递归函数，每一次递归都是一次每一位数字的置换，求出每一位的循环节，最终f(m)=k就是要每一位都相等，即解同余方程组。

代码：

#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include<climits>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <fstream>#include <numeric>#include <iomanip>#include <bitset>#include <list>#include <stdexcept>#include <functional>#include <utility>#include <ctime>using namespace std;#define PB push_back#define MP make_pair#define REP(i,x,n) for(int i=x;i<(n);++i)#define FOR(i,l,h) for(int i=(l);i<=(h);++i)#define FORD(i,h,l) for(int i=(h);i>=(l);--i)#define SZ(X) ((int)(X).size())#define ALL(X) (X).begin(), (X).end()#define RI(X) scanf("%d", &(X))#define RII(X, Y) scanf("%d%d", &(X), &(Y))#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))#define DRI(X) int (X); scanf("%d", &X)#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)#define OI(X) printf("%d",X);#define RS(X) scanf("%s", (X))#define MS0(X) memset((X), 0, sizeof((X)))#define MS1(X) memset((X), -1, sizeof((X)))#define LEN(X) strlen(X)#define F first#define S second#define Swap(a, b) (a ^= b, b ^= a, a ^= b)#define Dpoint  strcut node{int x,y}#define cmpd int cmp(const int &a,const int &b){return a>b;} /*#ifdef HOME    freopen("in.txt","r",stdin);    #endif*/const int MOD = 1e9+7;typedef vector<int> VI;typedef vector<string> VS;typedef vector<double> VD;typedef long long LL;typedef pair<int,int> PII;//#define HOMEint Scan(){int res = 0, ch, flag = 0;if((ch = getchar()) == '-')//判断正负flag = 1;else if(ch >= '0' && ch <= '9')//得到完整的数res = ch - '0';while((ch = getchar()) >= '0' && ch <= '9' )res = res * 10 + ch - '0';return flag ? -res : res;}/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <algorithm>using namespace std;const int MAXN = 410;struct bign{int len, s[MAXN];bign (){memset(s, 0, sizeof(s));len = 1;}bign (int num) { *this = num; }bign (const char *num) { *this = num; }bign operator = (const int num){char s[MAXN];sprintf(s, "%d", num);*this = s;return *this;}bign operator = (const char *num){for(int i = 0; num[i] == '0'; num++) ;  //去前导0len = strlen(num);for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';return *this;}bign operator + (const bign &b) const //+{bign c;c.len = 0;for(int i = 0, g = 0; g || i < max(len, b.len); i++){int x = g;if(i < len) x += s[i];if(i < b.len) x += b.s[i];c.s[c.len++] = x % 10;g = x / 10;}return c;}bign operator += (const bign &b){*this = *this + b;return *this;}void clean(){while(len > 1 && !s[len-1]) len--;}bign operator * (const bign &b) //*{bign c;c.len = len + b.len;for(int i = 0; i < len; i++){for(int j = 0; j < b.len; j++){c.s[i+j] += s[i] * b.s[j];}}for(int i = 0; i < c.len; i++){c.s[i+1] += c.s[i]/10;c.s[i] %= 10;}c.clean();return c;}bign operator *= (const bign &b){*this = *this * b;return *this;}bign operator - (const bign &b){bign c;c.len = 0;for(int i = 0, g = 0; i < len; i++){int x = s[i] - g;if(i < b.len) x -= b.s[i];if(x >= 0) g = 0;else{g = 1;x += 10;}c.s[c.len++] = x;}c.clean();return c;}bign operator -= (const bign &b){*this = *this - b;return *this;}bign operator / (const bign &b){bign c, f = 0;for(int i = len-1; i >= 0; i--){f = f*10;f.s[0] = s[i];while(f >= b){f -= b;c.s[i]++;}}c.len = len;c.clean();return c;}bign operator /= (const bign &b){*this  = *this / b;return *this;}bign operator % (const bign &b){bign r = *this / b;r = *this - r*b;return r;}bign operator %= (const bign &b){*this = *this % b;return *this;}bool operator < (const bign &b){if(len != b.len) return len < b.len;for(int i = len-1; i >= 0; i--){if(s[i] != b.s[i]) return s[i] < b.s[i];}return false;}bool operator > (const bign &b){if(len != b.len) return len > b.len;for(int i = len-1; i >= 0; i--){if(s[i] != b.s[i]) return s[i] > b.s[i];}return false;}bool operator == (const bign &b){return !(*this > b) && !(*this < b);}bool operator != (const bign &b){return !(*this == b);}bool operator <= (const bign &b){return *this < b || *this == b;}bool operator >= (const bign &b){return *this > b || *this == b;}string str() const{string res = "";for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;return res;}};istream& operator >> (istream &in, bign &x){string s;in >> s;x = s.c_str();return in;}ostream& operator << (ostream &out, const bign &x){out << x.str();return out;}int a[105],b[105];int ansm[1000],ansk[1000];int vis[105];int vis2[105];void ex_gcd(LL a, LL b, LL &d, LL &x, LL &y){    if (!b) {d = a, x = 1, y = 0;}    else    {        ex_gcd(b, a % b, d, y, x);        y -= x * (a / b);    }}LL ex_crt(LL *m, LL *r, int n,int lena,int cato){    LL M = lena, R = cato, x, y, d;    for (int i = 1; i < n; ++i)    {        ex_gcd(M, m[i], d, x, y);        if ((r[i] - R) % d) return -1;        x = (r[i] - R) / d * x % (m[i] / d);        R += x * M;        M = M / d * m[i];        R %= M;    }    return R > 0 ? R : R + M;}int main(){int d;while(RI(d)!=EOF){    if(d==-1)        break;        MS0(a);        MS0(b);        MS0(ansm);        MS0(ansk);    REP(i,1,d)    RI(a[i]);    REP(i,0,d)    RI(b[i]);    bign m,k;    cin>>m;    cin>>k;    bign tmp(0);    tmp.len=1;    bign r;    int p=0;    MS0(ansm);    MS0(ansk);    //cout<<m<<k<<endl;    while(m!=tmp)    {//cout<<m<<endl;    r=m%bign(d);    m=m/bign(d);    for(int i=r.len-1;i>=0;i--)        ansm[p]=ansm[p]*10+r.s[i];    p++;    }    for(int i=0;i<p/2;i++)        swap(ansm[i],ansm[p-1-i]);    int q=0;    while(k!=tmp)    {    r=k%bign(d);    k=k/bign(d);    for(int i=r.len-1;i>=0;i--)        ansk[q]=ansk[q]*10+r.s[i];    q++;    }    for(int i=0;i<q/2;i++)        swap(ansk[i],ansk[q-1-i]);    if(p!=q)    {        printf("NO\n");        continue;    }    MS0(vis);    int lena=0;    int now=ansm[0];    int ok=0;    int cato=0;    if(now==ansk[0])        ok=1;    vis[now]=1;    lena++;    while(!vis[a[now]])    {        lena++;        now=a[now];        vis[now]=1;        if(!ok)            cato++;        if(now==ansk[0])            ok=1;    }    long long int lenb[105];    long long int cto[105];    MS0(vis2);    int okb[105];    for(int i=1;i<p;i++)    {   MS0(vis2);        lenb[i]=0;        now=ansm[i];        okb[i]=0;        cto[i]=0;        if(now==ansk[i])            okb[i]=1;        vis2[now]=1;        lenb[i]++;        while(!vis2[b[now]])        {lenb[i]++;        now=b[now];        vis2[now]=1;        if(!okb[i])            cto[i]++;        if(now==ansk[i])            okb[i]=1;        }    }    if(!ok)    {        printf("NO\n");        continue;    }    int tok=1;    REP(i,1,p)    if(!okb[i])    {        tok=0;        printf("NO\n");        break;    }    if(!tok)        continue;    long long int res=ex_crt(lenb,cto,p,lena,cato);    if(res==-1)    {        printf("NO\n");        continue;    }    printf("%I64d\n",res);}        return 0;}