数学进位制相关模版

Posted 2022-11-04 legend_PawN

快速加模版

long long quickplus(long long  m,long long n,long long k)//返回m*n%k

    long long b = 0;
    while (n > 0)
    
          if (n & 1)
             b = (b+m)%k;
          n = n >> 1 ;
          m = (m+m)%k;
    
    return b;

快速幂取模模版:

typedef long long ll;
ll quickpow(ll a,ll b,ll c)
    int res=1;
    while(b>0)
        if(b&1)
            res=res*a%c;
        b=b>>1;
        a=(a%c)*(a%c)%c;
    
    return res;

矩阵快速幂模版:

const int N=4; //矩阵维数
struct mat
    int m[N][N];
    mat() 
    mat unit()   //单位矩阵 
        for(int i=0;i<N;i++)
            for(int j=0;j<N;j++)
                m[i][j]=i==j?1:0;
    
;
mat t; //t为状态转移矩阵 
mat operator * (mat a,mat b)  //方阵乘法 
    mat res;
    for(int i=0;i<N;i++)
        for(int j=0;j<=N;j++)
            res.m[i][j]=0;
            for(int k=0;k<N;k++)
                res.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;
        
    return res;

mat operator ^ (mat res,int n) //方阵二分幂 
    res.unit();
    while(n>=1)
        if(n&1)
            res=res*t;
        n=n>>1;
        t=t*t;
    
    return res;

void init() //初始化矩阵
 

排列组合数模版:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6+7;
const int mod = 1e9+7;
ll fac[maxn];
ll quickpow(ll a,ll b) //快速幂
    int res=1;
    while(b>0)
        if(b&1)
            res=res*a%mod;
        b=b>>1;
        a=(a%mod)*(a%mod)%mod;
    
    return res;

ll C(ll n,ll m)  //计算C(n,m)

    if(m>n||m<0)
        return 0;
    ll s1=fac[n],s2=fac[n-m]*fac[m]%mod;
    return s1*quickpow(s2,mod-2)%mod;

int main()

    fac[0]=1;             //预处理前缀
    for(int i=1;i<maxn;i++)
        fac[i]=fac[i-1]*i%mod;
    //printf("%lld\\n",C(4,3));

统计二进制中有多少个1:

popcount(int x)
    return cnt[x>>16]+cnt[x&((1<<16)-1)];