7.22 校内模拟赛

Posted 2020-09-29 ZlycerQan

 

 

题面和数据 : 二次联通门

 

 难度 : 普及-

 

 

 

/*
    T1 
    
    求多项式结果的后k位，由于k<=8， 
    所以不必考虑高精了，直接把多项式一遍算，
    一边取模，注意用快速幂，最后补0输出即可。
    复杂度 O (n log b)
*/
#include <cstdio>

#define Max 100090

void read (int &now)
{
    register char word = getchar ();
    bool temp = false;
    for (now = 0; word < \'0\' || word > \'9\'; word = getchar ())
        if (word == \'-\')
            temp = true;
    for (; word >= \'0\' && word <= \'9\'; now = now * 10 + word - \'0\', word = getchar ());
    if (temp)
        now = -now;
}

int Mod = 1;
int N, K, X;

int a[Max], b[Max];

long long Fast_Pow (long long x, long long p)
{
    register long long res = 1;
    for (; p; p >>= 1)
    {
        if (p & 1)
            res = res * x % Mod;
        x = x % Mod * x % Mod;
    }
    return res;
}

#define Judge
long long Answer;

int number[Max / 1000];

int main (int argc, char *argv[])
{

#ifdef Judge

    freopen ("digits.in", "r", stdin);
    freopen ("digits.out", "w", stdout);

#endif 

    read (N);
    read (K);

    for (int i = 1; i <= N; i ++)
    {
        read (a[i]);
        read (b[i]);
    }
    
    for (int i = K, res = 10; i; i >>= 1)
    {
        if (i & 1)
            Mod = Mod * res;
        res *= res;
    }
    read (X);
    
    for (int i = 1; i <= N; i ++)
        Answer = (Answer + (a[i] * Fast_Pow (X, b[i]) % Mod)) % Mod; 
    
    int Count = 0;
    for (; Answer > 0; number[++ Count] = Answer % 10, Answer /= 10);
    
    for (int i = Count; i < K; i ++, printf ("0\\n"));
    
    for (; Count >= 1; printf ("%d\\n", number[Count --]));
    return 0;
} 

 

 

 

 

 

 

 

 

/*
    T2
    
    把多项式左边的负项移到右边， 
    O(N^3)预处理出右边的结果，用一个桶记录下来，
    后O(N^3)枚举左边的结果， 加减就好了 
    复杂度O(N^3)
*/
#include <cstdio>

#define Max 60000002

#define N 6

void read (int &now)
{
    register char word = getchar ();
    bool temp = false;
    for (now = 0; word < \'0\' || word > \'9\'; word = getchar ())
        if (word == \'-\')
            temp = true;
    for (; word >= \'0\' && word <= \'9\'; now = now * 10 + word - \'0\', word = getchar ());
    if (temp)
        now = -now;
}

#define Judge

int K;

short z_count[Max];
short f_count[Max];


int a1, a2, a3, a4, a5, a6;

int main (int argc, char *argv[])
{

#ifdef Judge 
    
    freopen ("equation.in", "r", stdin);
    freopen ("equation.out", "w", stdout);
    
#endif
    
    read (K);
    read (a1);
    read (a2);
    read (a3);
    read (a4);
    read (a5);
    read (a6);    
    
    register int x;
    
    for (register int i = 1, j, k; i <= K; i ++)
        for (j = 1; j <= K; j ++)
            for (k = 1; k <= K; k ++)
            {
                x = a2 * i + a4 * j + a6 * k;
                if (x >= 0)
                    z_count[x] ++;
                else
                    f_count[-x] ++;
            }
            
    int Answer = 0;
    
    for (register int i = 1, j, k; i <= K; i ++)
        for (j = 1; j <= K; j ++)
            for (k = 1; k <= K; k ++)
            {
                x = a1 * i + a3 * j + a5 * k;
                if (x >= 0)
                    Answer += z_count[x];
                else
                    Answer += f_count[-x];
            }
    
    printf ("%d", Answer);
    return 0;
}

 

 

 

 

/*
    T3 
    
    由于k>=n，所以一天最多走一条边，
    则问题转化为了求最小瓶颈生成树的裸题。。。。
    若一直到最后起点与终点都不联通，那么则无解
    
    复杂度 O(M logM + M) = O (M log M)
*/
#include <algorithm>
#include <cstdio>

#define Max 8000

void read (int &now)
{
    register char word = getchar ();
    for (now = 0; word < \'0\' || word > \'9\'; word = getchar ());
    for (; word >= \'0\' && word <= \'9\'; now = now * 10 + word - \'0\', word = getchar ());
}

inline int max (int a, int b)
{
    return a > b ? a : b;
}

struct Edge
{
    int from;
    int to;
    int dis;

    bool operator < (const Edge &now) const
    {
        return this->dis < now.dis;
    }
};

class Unio_Find_Set
{
    
    private :
        
        int father[Max];
        
    public :

        void Prepare (int N)
        {
            for (int i = 1; i <= N; i ++)
                father[i] = i;
        }
        
        int Find (int x)
        {
            return father[x] == x ? x : father[x] = Find (father[x]);
        }

        inline void Unio (int a, int b)
        {
            father[a] = b;
        }
};

Unio_Find_Set Ufs;

int N, M, K;

Edge edge[Max * 30];
#define Judge

int main (int argc, char *argv[])
{

#ifdef Judge 
    
    freopen ("graph.in", "r", stdin);
    freopen ("graph.out", "w", stdout);
    
#endif
    
    read (N);
    read (M);
    read (K);
    
    for (int i = 1; i <= M; i ++)
    {
        read (edge[i].from);
        read (edge[i].to);
        read (edge[i].dis);
    }
    
    Ufs.Prepare (N);
    std :: sort (edge + 1, edge + 1 + M);
    
    int Count = 0;
    for (register int i = 1, x, y; i <= M; i ++)
    {
        x = Ufs.Find (edge[i].from);
        y = Ufs.Find (edge[i].to);
        
        if (x != y)
        {
            Ufs.Unio (x, y); 
            Count ++;
        }
        if (Ufs.Find (1) == Ufs.Find (N))
        {
            printf ("%d", edge[i].dis);
            return 0;
        }
        if (Count == N - 1)
            break;
    }
    printf ("-1");
    return 0;
}