竞赛大纲

#include <cstdio>
#include <cmath>
#include <queue>
#include <iostream>
#include <algorithm>

//一条路可以重复走多次
using namespace std;
const int N = 200070, inf = 0x3f3f3f3f;
struct Edg {
    int to, val, nxt;
} e[N];
int n, m, cnt, d[N], D[N], head[N];
priority_queue <  pair<int, int> > q;
void add(int from, int to, int val){
    e[++cnt].to = to;
    e[cnt].val = val;
    e[cnt].nxt = head[from];
    head[from] = cnt;
}
void dij(){
    for (int i = 1; i <= n; ++i) 
        d[i] = D[i] = inf;
    d[1] = 0;
    q.push( make_pair(1,0) );
    while( !q.empty() ){
        int u = q.top().first, frontdis = -q.top().second; q.pop();
        for (int i = head[u]; i; i = e[i].nxt) {
            int v = e[i].to, dis = frontdis + e[i].val;
            if ( d[v] > dis ) {
                D[v] = d[v], d[v] = dis;
                q.push( make_pair(v, -d[v]) );
            }
            if ( D[v] > dis && d[v] < dis ){
                D[v] = dis;
                q.push( make_pair(v, -D[v]) );
            }
        }
    }
}
int main(){
    scanf("%d %d", &n, &m);
    for (int i = 1, u, v, w; i <= m; ++i) {
        scanf("%d %d %d", &u, &v, &w);
        add(u, v, w), add(v, u ,w);
    }
    dij();
    printf("%d
", D[n]);
    return 0;
}

#include <cstdio>
#include <algorithm>

using namespace std;
const int N = 1e6 + 70;
int n, cnt, phi[N], prime[N], vis[N];
int main(){
    phi[1] = 1;
    for (int i = 2; i <= N; ++i) {
        if ( !vis[i] ) {
            prime[++cnt] = i;
            phi[i] = i - 1;
        }
        // phi[ a * b ] = phi[a] * phi[b];
        for (int j = 1; j <= cnt && i * prime[j] <= N; ++j){
            vis[ i * prime[j] ] = 1;
            phi[ i * prime[j] ] = i%prime[j] ? (prime[j] - 1) * phi[i] :
                                  prime[j] * phi[i]; 
        }
    }
    int T, n;
    scanf("%d", &T);
    while ( T-- ) {
        scanf("%d", &n);
        printf("%d
", phi[n]);
    }
    return 0;
}

inv[N] = qpow(f[N], p - 2);
    for (int i = N - 1; i >= 0; --i) 
        inv[i] = 1LL * inv[i + 1] % p * (i + 1) % p;

#include <cstdio>
#include <cmath>

using namespace std;
int gcd(int a, int b){
    return b ? gcd(b, a % b) : a;
}
int main(){
    int n, ans = 0;
    scanf("%d", &n);
    for (int i = 1, x; i <= n; ++i) {
        scanf("%d", &x);
        ans = gcd(ans, abs(x));
    }
    printf("%d
", ans);
    return 0;
}

inv[1] = 1;
    for (int i = 2; i <= n; ++i) 
        inv[i] = (p - p/i) * inv[p % i] % p;

错排公式

D[0] = 0, D[1] = 0, D[2] = 1;
for (int i = 3; i <= N; ++i) 
        D[i] = 1LL * (i - 1) * (D[i - 1] + D[i - 2]) % p;