蓝桥杯百校真题大联赛第3期（三）

A组

RSA 解密：扩展欧几里得、快速幂

#include<bits/stdc++.h>

using namespace std;
typedef __int128 Int;
typedef long long ll;
        ll n = 1001733993063167141, d = 212353 ,C = 20190324,tmp;
        Int e;
bool isPrime(int x){
    if(x<2) return false;
    for(int i=2;i<=sqrt(x);i++)
        if(x%i==0) return false;
    return true;
}

void Print(Int x){
    int a[1000],top=0;
    while(x){
        int b=x%10;
        a[++top]=b;
        x/=10;
    }
    for(int i=top;i>=1;i--)
    cout<<a[i];
}

void qmod(ll a,ll n,ll mod){
    Int res=1;
    Int base=a;
        while(n){
            if(n&1) res=(res*base)%mod;
            base=(base*base)%mod;
            n>>=1;
        }
    Print(res);
    
}

int main(){
    ll p,q;
    //1，求p,q; 
//    for(int i=3;i<=sqrt(n);i++)
//    if(n%i==0&&isPrime(i)&&isPrime(n/i)){
//        p=i;q=n/i;
//        break;
//    }
//cout<<p<<" "<<q;
    p=891234941; q=1123984201;

//2，求e
    tmp=(p-1)*(q-1);
    for(Int i=2;i<=n;i++){
        Int now=i*tmp+1;
        if(now%d==0){
            e=now/d;
            break;
        }
    } 
//Print(e);

    qmod(C,e,n);
    return 0;
}

完全二叉树的权值：堆

#include <iostream>
#include <cstring>
#include <algorithm>

typedef long long LL;

using namespace std;
const int N = 100010;
int n;
int a[N];

int main()
{
    LL sum = -0x3f3f3f3f, depth = 1;
    scanf("%d", &n);
    
    for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
    
    for (int d = 1, i = 1; i <= n; i ++ , d ++ )
    {
        LL j = i, s = 0;
        while (j < i * 2 && j <= n)
        {
            s += a[j];
            j ++ ;
        }
        
        if (s > sum)
        {
            sum = s;
            depth = d;
        }
        i = j - 1;
    }
    cout << depth << endl;
    return 0;
}

B组

七段码：并查集、连通块、枚举

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 10;
int n;
int g[N][N];
bool st[N];

void dfs(int t, int u)
{
    st[u] = true;
    for (int j = 0; j < 7; j ++ )
        if (g[u][j + 1] && (t >> j & 1) && !st[j + 1])
            dfs(t, j + 1);
}

bool check(int x)
{
    int t = x, cnt = 0;
    memset(st, 0, sizeof st);
    for (int j = 0; j < 7; j ++ )
        if (t >> j & 1 && !st[j + 1])
        {
            dfs(t, j + 1);
            cnt ++ ;
        }
    return cnt == 1;
}

int main()
{
    for (int i = 1; i <= 6; i ++ )
    {
        g[i][i + 1] = g[i + 1][i] = 1;
    }
    
    g[2][7] = g[7][2] = 1;
    g[3][7] = g[7][3] = 1;
    g[5][7] = g[7][5] = 1;
    g[1][6] = g[6][1] = 1;
    
    int ans = 0;
    for (int i = 1; i < (1 << 7); i += 1)
    {
        if (check(i)) ans ++ ;
    }
    
    cout << ans << endl;
    
    return 0;
}

成绩统计：格式化输出

#include <iostream>

using namespace std;

int main()
{
    int t,a,n;
    double j = 0,y = 0;
    cin>>t;
    n = t;
    while(t--){
        cin >> a;
        if(a>=60) j++;
        if(a>=85) y++;
    }
    printf("%.0f%%\n",(j/n)*100); 
    printf("%.0f%%\n",(y/n)*100);
    return 0;
}

C组

路径

朴素版dijkstra算法

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 2200;

int g[N][N];
int dist[N];
int n = 2021;
bool st[N];

int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}

void dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    
    for (int i = 1; i <= n; i ++ )
    {
        int t = -1;
        for (int j = 1; j <= n; j ++ )
            if (!st[j] && (t == -1 || dist[j] < dist[t]))
                t = j;
        
        st[t] = true;
        for (int j = 1; j <= n; j ++ )
            dist[j] = min(dist[j], dist[t] + g[t][j]);
    }
}

int main()
{
    memset(g, 0x3f, sizeof g);
    for (int i = 1; i <= 2021; i ++ )
        for (int j = max(i - 21, 1); j <= min(i + 21, n); j ++ )
        {
            int d = gcd(i, j);
            g[i][j] = g[j][i] = i * j / d;
        }
    
    dijkstra();
    
    cout << dist[n] << endl;
    
    return 0;
}

堆优化版dijkstra算法

在这里插入代码片

spfa算法

在这里插入代码片

时间显示

#include <iostream>
#include <cstring>
#include <algorithm>
typedef long long LL;

using namespace std;

int main()
{
    LL n, h, m, s;
    cin >> n;
    
    n /= 1000;
    h = n / 3600 % 24;
    
    n %= 3600;
    s = n % 60;
    m = n / 60;
    
    printf("%02lld:%02lld:%02lld", h, m, s);
    return 0;
}