URAL1013 K-based Numbers. Version 3(矩阵快速幂+大数)

题目描述：

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题意要求你组成一个N位的K进制数要求没有前导0和相邻的0，求所有能组成的方案总数.在K一定的情况下去考虑N，N=1,2,3时都非常好算，从N等于4开始，由于最高位不能是0，所以最高位只有K-1种选法，先在前K -1位的基础上用(K-1)*f[K-1](f[x]表示x位的方案总数)，那么还多出来的方案数就是K-1位上选0的情况，由于不能有相邻的0，那么此时K-2位上就不能选0，那么这些方案总数其实就是f[K-2]，所以递推式就是f[x]=(K-1)*(f[x-1]+f[x-2])(x>=4)，对于这个式子用矩阵快速幂去跑就好了，另外由于这个题数据非常变态，所以所有的数据全部都得用大数。

AC代码：

#include<iostream>#include<time.h>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<stack>#include<queue>#include<vector>#include<set>#include<algorithm>using namespace std;const int SIZE=50;const int INF=0x3f3f3f3f;const int MAXM=2000;struct BigInteger {    int len, s[SIZE + 5];    BigInteger () {        memset(s, 0, sizeof(s));        len = 1;    }    BigInteger operator = (const char *num) { //字符串赋值        memset(s, 0, sizeof(s));        len = strlen(num);        for(int i = 0; i < len; i++) s[i] = num[len - i - 1] - '0';        return *this;    }    BigInteger operator = (const long long num) { //int 赋值        memset(s, 0, sizeof(s));        char ss[SIZE + 5];        sprintf(ss, "%lld", num);        *this = ss;        return *this;    }    BigInteger (long long num) {        *this = num;    }    BigInteger (char* num) {        *this = num;    }    string str() const { //转化成string        string res = "";        for(int i = 0; i < len; i++) res = (char)(s[i] + '0') + res;        if(res == "") res = "0";        return res;    }    BigInteger clean() {        while(len > 1 && !s[len - 1]) len--;        return *this;    }    BigInteger operator + (const BigInteger& b) const {        BigInteger 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.clean();    }    BigInteger operator - (const BigInteger& b) {        BigInteger 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;        }        return c.clean();    }    BigInteger operator * (const int num) const {        int c = 0, t;        BigInteger pro;        for(int i = 0; i < len; ++i) {            t = s[i] * num + c;            pro.s[i] = t % 10;            c = t / 10;        }        pro.len = len;        while(c != 0) {            pro.s[pro.len++] = c % 10;            c /= 10;        }        return pro.clean();    }    BigInteger operator * (const BigInteger& b) const {        BigInteger c;        for(int i = 0; i < len; i++) {            for(int j = 0; j < b.len; j++) {                c.s[i + j] += s[i] * b.s[j];                c.s[i + j + 1] += c.s[i + j] / 10;                c.s[i + j] %= 10;            }        }        c.len = len + b.len + 1;        return c.clean();    }    BigInteger operator / (const BigInteger &b) const {        BigInteger c, f;        for(int i = len - 1; i >= 0; --i) {            f = f * 10;            f.s[0] = s[i];            while(f >= b) {                f = f - b;                ++c.s[i];            }        }        c.len = len;        return c.clean();    }    //高精度取模    BigInteger operator % (const BigInteger &b) const{        BigInteger r;        for(int i = len - 1; i >= 0; --i) {            r = r * 10;            r.s[0] = s[i];            while(r >= b) r = r - b;        }        r.len = len;        return r.clean();    }    bool operator < (const BigInteger& b) const {        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 BigInteger& b) const {        return b < *this;    }    bool operator <= (const BigInteger& b) const {        return !(b < *this);    }    bool operator == (const BigInteger& b) const {        return !(b < *this) && !(*this < b);    }    bool operator != (const BigInteger &b) const {        return !(*this == b);    }    bool operator >= (const BigInteger &b) const {        return *this > b || *this == b;    }    friend istream & operator >> (istream &in, BigInteger& x) {        string s;        in >> s;        x = s.c_str();        return in;    }    friend ostream & operator << (ostream &out, const BigInteger& x) {        out << x.str();        return out;    }};struct MAT{    BigInteger s[2][2];    MAT () {        memset(s,0,sizeof(s));    }};long long n,m,k;MAT MAT_MUL(MAT &m1,MAT &m2){    MAT m3;    for (int i=0;i<2;i++) {        for (int j=0;j<2;j++) {            for (int k=0;k<2;k++)                m3.s[i][j]=(m3.s[i][j]+m1.s[i][k]*m2.s[k][j])%m;        }    }    return m3;}int main(){    while(scanf("%lld%lld%lld",&n,&k,&m)!=EOF)    {        BigInteger K(k),M(m);        BigInteger ans;        if (n==1) {            ans=K-1;            cout<<ans<<endl;            continue;        }        else if (n==2) {            ans=(K%M*(K-1+M)%M)%M;            cout<<ans<<endl;            continue;        }        else if (n==3) {            BigInteger tmp=(K%M*(K-1+M)%M)%M;            tmp=(tmp+K-1+M)%M;            ans=((K-1+M)%M*tmp)%M;            cout<<ans<<endl;            continue;        }        BigInteger f2=(K%M*(K-1+M)%M)%M;        BigInteger tmp=(K%M*(K-1+M)%M)%M;        tmp=(tmp+K-1+M)%M;        BigInteger f3=((K-1+M)%M*tmp)%M;        //cout<<f2<<' '<<f3<<endl;        K=K-1;        MAT m1;        m1.s[0][0]=f3;m1.s[0][1]=(long long)0;        m1.s[1][0]=f2;m1.s[1][1]=(long long)0;        MAT m2;        m2.s[0][0]=K;m2.s[0][1]=K;        m2.s[1][0]=(long long)1;m2.s[1][1]=(long long)0;        n=n-3;        while(n) {            if (n&1) m1=MAT_MUL(m2,m1);            m2=MAT_MUL(m2,m2);            n=n>>1;        }        ans=m1.s[0][0];        /*for (int i=0;i<2;i++) {            for (int j=0;j<2;j++)                printf("%lld ",m1.s[i][j]);            printf("\n");        }*/        cout<<ans<<endl;    }    return 0;}