模板区

这是用来存模板的地方，不断更新。 

基础

输入优化（整数）

inline int read(){
    int num = 0, b = 1;
    char c = getchar();
    while(c < '0' || c > '9') c == '-' ? b = -1 : 0, c = getchar();
    while(c >= '0' & c <= '9') num = num * 10 + c - '0', c = getchar();
    return num * b;
}

数学

最大公约数

int gcd(int a, int b){
    if(!b) return a;
    return gcd(b, a % b);
}

快速幂

int qpow(int a, int b){                       // 分别代表底数、指数
    int t = 1;
    while(b){
        if(b & 1) t = t * a % MOD;       　　  // MOD 是模数
        a = a * a % MOD;
        b >>= 1;
    }
    return t;
}

埃氏筛法

bool b[1000010];            　　　　　　　　　　// b[i] 代表 i 是否是质数
void Prime(){
    for(int i = 2; i <= 1000; i++){
        if(!b[i])
            for(int j = i; j <= 1000000; j += i)
                b[j] = 1;
    }
}

线性筛

int p[1000010], ans[1000010], cnt;                // 存储质数的序列
void Prime(){
    for(int i = 2; i <= 1000000; i++){
        if(!b[i]) ans[++cnt] = i;
        for(int j = 1; i * p[j] <= 1000000; j++){
            b[i % p[j]] = 1;
            if(i % p[j] == 0) break;
         }
    }
}

求单个数在模数为质数意义下的逆元

typedef long long ll;
ll qpow(ll a, ll b){                       　　// 分别代表底数、指数
    ll t = 1;
    while(b){
        if(b & 1) t = t * a % MOD;       　　  // MOD 是模数
        a = a * a % MOD;
        b >>= 1;
    }
    return t;
}
ll inv(ll x){
    return qpow(x, MOD - 2);
}

线性求范围内模数为质数意义下的逆元

typedef long long ll;
ll inv[1000010];
void Inv(){
    inv[0] = inv[1] = 0;
    for(ll i = 2; i <= 1000000; i++)
        inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;    // MOD 是模数
}

数字 $aa$ 在模 $bb$ 意义下（不保证是质数）的逆元

typedef long long ll;
void exgcd(ll a, ll b, int &x, int &y){
    if(!b){
        x = 1, y = 0;
        return
    }
    exgcd(b, a % b, x, y);
    y -= (a / b) * x;
}
ll inv(ll aa, ll bb){
    ll x = 0, y = 0;
    exgcd(aa, bb, x, y);
    return (x + bb) % bb;    
}

图论

最小生成树（竞速：285s）

#include <algorithm>
#include <cstdio>
#define INF 1e9
#define eps 1e-9
typedef long long ll;

struct Edge{
    int from;
    int to;
    int val;
}e[400010];
int n, m, f[10010], x, y, z, cnt, sum, ans;

void add(int a, int b, int c){
    e[++cnt].from = a;
    e[cnt].to = b;
    e[cnt].val = c;
}

bool cmp(Edge a, Edge b){
    return a.val < b.val;
}

int find(int xx){
    if(xx == f[xx]) return xx;
    return f[xx] = find(f[xx]);
}

int main(){
    
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++)
        scanf("%d%d%d", &x, &y, &z), add(x, y, z), add(y, x, z);
    std::sort(e + 1, e + cnt + 1, cmp);
    for(int i = 1; i <= n; i++) f[i] = i;
    for(int i = 1; i <= cnt; i++){
        x = find(e[i].from), y = find(e[i].to);
        if(x != y){
            f[x] = y, ans += e[i].val;
            if(++sum == n - 1) break;
        }
    }
    if(sum != n - 1) puts("orz");
    else printf("%d
", ans);


    return 0;
}

 单源最短路径（Dijkstra）（竞速：482s）

#include <cstring>
#include <cstdio>
#include <queue>
#define INF (1 << 30)
#define eps 1e-9
#define N 100010
typedef long long ll;
using namespace std;

struct Edge{
    int to, next, val;
}e[N << 1]; 

struct Point{
    int id, dis;
};

bool operator < (Point x, Point y){
    return x.dis > y.dis;
}
 
priority_queue <Point> q;
int n, m, s, fir[N], u, v, w, cnt, dis[N];
bool vis[N];

void add(int x, int y, int z){
    e[++cnt].to = y;
    e[cnt].next = fir[x];
    e[cnt].val = z;
    fir[x] = cnt;
}

void Dijkstra(){
    for(int i = 1; i <= n; i++)
        dis[i] = INF;
    dis[s] = 0;
    q.push((Point){s, 0});
    while(!q.empty()){
        int x = q.top().id;
        q.pop();
        if(vis[x]) continue;
        vis[x] = 1;
        for(int i = fir[x]; i; i = e[i].next){
            int y = e[i].to;
            if(dis[y] > dis[x] + e[i].val){
                dis[y] = dis[x] + e[i].val;
                q.push((Point){y, dis[y]});
            }
        }
    }
    for(int i = 1; i <= n; i++)
        printf("%d ", dis[i]);
    puts("");
}

int main(){
    
    scanf("%d%d%d", &n, &m, &s);
    for(int i = 1; i <= m; i++)
        scanf("%d%d%d", &u, &v, &w), add(u, v, w); 
    Dijkstra();

    return 0;
}

数据结构

线段树（区间加，区间和）（竞速：450s）

#include <cstdio>
#define N 100010
#define INF 1e9
#define eps 1e-9
typedef long long ll;

int n, m, p, q, opt;
ll s, a[N << 2], add[N << 2], sum[N << 2];

void build(int i, int l, int r){
    if(l == r){
        sum[i] = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(i << 1, l, mid);
    build(i << 1 | 1, mid + 1, r);
    sum[i] = sum[i << 1] + sum[i << 1 | 1];
}

void plu(int i, int l, int r, ll v){
    add[i] += v;
    sum[i] += (r - l + 1) * v;
}

void pushdown(int i, int l, int r, int mid){
    if(add[i] == 0) return;
    plu(i << 1, l, mid, add[i]);
    plu(i << 1 | 1, mid + 1, r, add[i]);
    add[i] = 0;
}

void modify(int i, int l, int r, int x, int y, ll v){
    if(l >= x && r <= y){
        plu(i, l, r, v);
        return;
    }
    int mid = (l + r) >> 1;
    pushdown(i, l, r, mid);
    if(x <= mid) modify(i << 1, l, mid, x, y, v);
    if(y > mid) modify(i << 1 | 1, mid + 1, r, x, y, v);
    sum[i] = sum[i << 1] + sum[i << 1 | 1];
}

ll query(int i, int l, int r, int x, int y){
    if(l >= x && r <= y) return sum[i];
    int mid = (l + r) >> 1;
    ll ans = 0;
    pushdown(i, l, r, mid);
    if(x <= mid) ans += query(i << 1, l, mid, x, y);
    if(y > mid) ans += query(i << 1 | 1, mid + 1, r, x, y);
    return ans;
}

int main(){
    
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++)
        scanf("%lld", &a[i]);
    build(1, 1, n);
    while(m--){
        scanf("%d%d%d", &opt, &p, &q);
        if(opt == 1)
            scanf("%lld", &s), modify(1, 1, n, p, q, s);
        else
            printf("%lld
", query(1, 1, n, p, q));
    }

    return 0;
}

 LCA（倍增）

#include <cstdio>
#define INF 1e9
#define eps 1e-9
#define N 500010
typedef long long ll;

struct Edge{
    int to, next;
}e[N << 1];
int n, m, s, u, v, lg[N];
int d[N], fir[N], cnt, f[N][25];

void add(int x, int y){
    e[++cnt].to = y;
    e[cnt].next = fir[x];
    fir[x] = cnt;
}

void dfs(int x, int fa){
    d[x] = d[fa] + 1, f[x][0] = fa;
    for(int i = 1; (1 << i) <= d[x]; i++)
        f[x][i] = f[f[x][i - 1]][i - 1];
    for(int i = fir[x]; i; i = e[i].next)
        if(e[i].to != fa)
            dfs(e[i].to, x);
}

int lca(int x, int y){
    if(d[x] < d[y]) x ^= y, y ^= x, x ^= y;
    while(d[x] > d[y])
        x = f[x][lg[d[x] - d[y]] - 1];
    if(x == y) return x;
    for(int i = lg[d[x]] - 1; i >= 0; i--)
        if(f[x][i] != f[y][i])
            x = f[x][i], y = f[y][i];
    return f[x][0];
}

int main(){
    
    scanf("%d%d%d", &n, &m, &s);
    for(int i = 1; i < n; i++)
        scanf("%d%d", &u, &v), add(u, v), add(v, u);
    dfs(s, 0);
    for(int i = 1; i <= n; i++)
        lg[i] = (1 << lg[i - 1]) == i ? lg[i - 1] + 1 : lg[i - 1];
    while(m--){
        scanf("%d%d", &u, &v);
        printf("%d
", lca(u, v));
    }

    return 0;
}

DP

……先咕着，以后再存