实现技巧

#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <string>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <complex>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
typedef vector<int> vi;
#define de(x) cout << #x << "=" << x << endl
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
const int N = 101010;

// Geometry
const db eps = 1e-8;
int sgn(db x){return (x > eps) - (x < -eps);}
typedef complex<db> P;
db dot(P a,P b){return (conj(a)*b).real();}
db cross(P a,P b){return (conj(a)*b).imag();}

// Matrix
typedef vector<vi> MM;
MM operator * (const MM a,const MM b){
    int L = sz(a);MM r(L,vi(L,0));
    rep(i,0,L) r[i][i] = 0;
    rep(i,0,L) rep(j,0,L) rep(k,0,L) r[i][j] += a[i][k] * b[k][j];
    return r;
}
MM operator ^ (MM a,ll t){
    MM r(sz(a),vi(sz(a),0));
    rep(i,0,sz(a)) r[i][i] = 1;
    for(;t;t>>=1,a=a*a) if(t&1) r=r*a;
    return r;
}

// Plus
const int MOD = 1e9 + 7;
void pp(int &x,int d){
    x += d;if(x >= MOD) x -= MOD;
}
// pp(a , P - x);

// treeDP
vi g[N];
int sz[N];
void dfs(int c,int pa){
    sz[c] = 1;
    for(auto t : g[c]) if(t != pa){ // c++11
        dfs(t , c);
        sz[c] += sz[t];
    }
}

// dsu
int fa[N];
int F(int x){return fa[x] == x ? x : fa[x] = F(fa[x]);}
void M(int x,int y){fa[F(x)]=F(y);}

// 两倍空间的线段树
#define id(l,r) (l+r)|(l!=r)
#define pr int c=id(l,r),m=(l+r)>>1
int sum[N<<1];
int Build(int l,int r){pr;
    if(l == r) return sum[c] = 1;
    else return sum[c] = Build(l,m) + Build(m+1,r);
}

int main(){
    int u = 0, v = 1;
    std::swap(u , v); // swap

    int a[20] , n = 20;
    cout << *std::max_element(a , a + n) << endl;// [a , a+n)
    cout << *std::min_element(a , a + n) << endl;

    vi V;// about 10 int
    sort(all(V));V.erase(unique(all(V),V.end())); // 离散化
#define Rk(x) lower_bound(all(V) , x) - V.begin() + 1

    for(int i=0,j=0;i<sz(V);i=j){ // 相同值处理
        for(j=i;j<sz(V)&&V[j]==V[i];++j);
        // Cal(i , j) //[i , j)
    }

    int g[10][10] , m = 10;
    rep(i,0,n) rep(j,0,m) scanf("%d",&g[i][j]); // 同逻辑共行

#define lb(x) (x&-(x))
    int cnt[1<<6];
    rep(i,1,1<<6) cnt[i] = cnt[i^lb(i)] + 1; // __builtin_popcount();

    sort(all(V),[&](int a,int b){return cnt[a]<cnt[b];}); // c++11

    rep(mask,0,1<<10) // 枚举子集
        for(int j=mask;j;j=(j-1)&mask)
            ;// Cal

    int f[1<<10];
    rep(i,0,10) rep(j,0,1<<10) if(j>>i&1) pp(f[j],f[j^(1<<i)]); // 高维前缀和

    vector<pii> p; // pair 排序
    rep(i,0,20) p.pb(mp(rand(),i));
    sort(all(p));

    rep(i,0,7) a[i] = i; // 枚举排列
    do{
        // Cal;
    }while(next_permutation(a , a + 7));

    fill(a , a + 20 , 0);// any number

    int &r=f[10];// 善用引用
    rep(i,0,10) r+=i;

    int C[10][10] = {{1}};// 适当的三目运算符
    rep(i,1,10) rep(j,0,i+1) C[i][j] = j ? (C[i-1][j-1] + C[i-1][j]) : 1;

    return 0;
}
原文地址:https://www.cnblogs.com/klaycf/p/9798467.html