有待整理的模板

矩阵快速幂

 

#include<iostream>
#include<cstring>
#include<queue>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define N 10
using namespace std;
struct Matrix{
    LL m[N][N];
}init,unit;
int MOD;
Matrix Mult(Matrix m1,Matrix m2,int n=2){
    Matrix ans;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++){
            ans.m[i][j]=0;
            for(int k=0;k<n;k++)
                ans.m[i][j]=((LL)ans.m[i][j]+m1.m[i][k]*m2.m[k][j])%MOD;
        }
    return ans;
}
Matrix Pow(Matrix m1,int b,int n=2){
    Matrix ans;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            ans.m[i][j]=(i==j);
    while(b){
        if(b&1)
            ans=Mult(ans,m1,n);
        m1=Mult(m1,m1,n);
        b>>=1;
    }
    return ans;
}
Matrix Add(Matrix m1,Matrix m2,int n=2){
    Matrix ans;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            ans.m[i][j]=((LL)m1.m[i][j]+m2.m[i][j])%MOD;
    return ans;
}
Matrix slove(Matrix init,int k,int n=2){
    if(k==1)
        return init;
    Matrix temp=slove(init,k>>1,n);
    temp=Add(temp,Mult(temp,Pow(init,k>>1)));
    if(k&1)
        return Add(temp,Pow(init,k));
    else
        return temp;
}
int main(){
    int k,b,n;
    while(scanf("%d%d%d%d",&k,&b,&n,&MOD)!=EOF){
        init.m[0][0]=init.m[0][1]=init.m[1][0]=1;
        init.m[1][1]=0;
        unit.m[0][0]=unit.m[1][1]=1;
        unit.m[1][0]=unit.m[0][1]=0;
        Matrix t;
        t=Pow(init,b);
        init=Pow(init,k);
        //init.m[0][1]即第k项fib，为A^k
        init=Add(unit,slove(init,n-1)); //B=A^k,   B^0+B^1+B^2……+B^n-1    
        init=Mult(init,t);
        printf("%I64d\n",init.m[0][1]);
    }
    return 0;
}

 快速幂

long long quick_mod(long long a,long long b,long long m) {
    long long ans = 1;
    while (b) {
        if (b&1) {
            ans = (ans * a) % m;
            b--;
        }
        b/=2;
        a = a * a% m;
    }
    return ans;
}

 组合数求模（逆元）

ll powmod(ll a, ll n)
{
    ll ret = 1;
    for (; n; n>>=1, a=a*a%mod)
        if (n&1)
        ret = ret*a%mod;
    return ret % mod;
}
void init(ll m, ll k)
{
    C[0] = 1;
    for (ll i=1;i<=k;i++)
        C[i] = ( C[i-1] * (m-i+1) %mod )* inv[i]% mod;
}

memset(inv, 0, sizeof inv);
    for (int i=1;i<N;i++) inv[i] = powmod(i, mod-2)%mod;

 

 

分数表示模板

 

#include <cstdio>
#include <cstring> #include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 500;

struct fraction {
    long long numerator; // 分子
    long long denominator; // 分母
    fraction() {
        numerator = 0;
        denominator = 1;
    }
    fraction(long long num) {
        numerator = num;
        denominator = 1;
    }
    fraction(long long a, long long b) {
        numerator = a;
        denominator = b;
        this->reduction();
    }

    void operator = (const long long num) {
        numerator = num;
        denominator = 1;
        this->reduction();
    }

    void operator = (const fraction &b) {
        numerator = b.numerator;
        denominator = b.denominator;
        this->reduction();
    }

    fraction operator + (const fraction &b) const {
        long long gcdnum = __gcd(denominator, b.denominator);
        fraction tmp = fraction(numerator*(b.denominator/gcdnum) + b.numerator*(denominator/gcdnum), denominator/gcdnum*b.denominator);
        tmp.reduction();
        return tmp;
    }

    fraction operator + (const int &b) const {
        return ((*this) + fraction(b));
    }

    fraction operator - (const fraction &b) const {
        return ((*this) + fraction(-b.numerator, b.denominator));
    }

    fraction operator - (const int &b) const {
        return ((*this) - fraction(b));
    }

    fraction operator * (const fraction &b) const {
        fraction tmp = fraction(numerator*b.numerator, denominator * b.denominator);
        tmp.reduction();
        return tmp;
    }

    fraction operator * (const int &b) const {
        return ((*this) * fraction(b));
    }

    fraction operator / (const fraction &b) const {
        return ((*this) * fraction(b.denominator, b.numerator));
    }

    void reduction() {
        if (numerator == 0) {
            denominator = 1;
            return;
        }
        long long gcdnum = __gcd(numerator, denominator);
        numerator /= gcdnum;
        denominator /= gcdnum;
    }
    void print() {
        if (denominator == 1) printf("%lld\n", numerator);
        else {
            long long num = numerator/denominator;
            long long tmp = num;
            int len = 0;
            while (tmp) {
                len++;
                tmp/=10;
            }

            for (int i = 0; i < len; i++) printf(" ");
            if (len != 0) printf(" ");
            printf("%lld\n",numerator%denominator);

            if (num != 0) printf("%lld ", num);
            tmp = denominator;
            while (tmp) {
                printf("-");
                tmp/=10;
            }
            puts("");

            for (int i = 0; i < len; i++) printf(" ");
            if (len != 0) printf(" ");
            printf("%lld\n",denominator);
        }
    }
} f[maxn];

int main() {
    int n;
    while (scanf("%d", &n) != EOF)
    {
        f[0] = 0;
        for (int i = 1; i <= n; i++)
            f[i] = f[i-1] + fraction(1, i);
        f[n] = f[n] * n;

        f[n].print();
    }
    return 0;
}