20171107校内模拟赛

 

 lqz已经什么题也写不对了。

 

期望得分：100 + 100 + 100

实际得分：95  + 70  + 60

 

 

T1本来写的矩阵快速幂，可是怎么也调不出来（初始矩阵建错），于是就去打了个找循环节的做法，自己手算了一下貌似循环节的长度大部分都很短，于是就去写了，但是被卡掉一个点。

 

T2打了个表，一下子发现是个等差数列，于是就写了正解，但是自己智障，特判了一下n==m的情况，这本来不用特判，正常跑就能跑出来，但是自己莫名其妙就认为当n==m的时候答案是固定的，于是挂了30分，去掉后就可以AC。

 

T3本来打的是标算，但是打错了一个字母，哭死。ev[i] -> _v[i]

 

T1

 

/*
    因为是mod7
    所以可以找一下从哪里开始循环之前的值
    找到就退出 
*/
#include <cstdio>
#include <iostream>

#define rg register
typedef long long LL;
#define Max 55000000
short f[Max];
std :: string Name = "attack", _I = ".in", _O = ".out";
int main (int argc, char *argv[])
{
    freopen ((Name + _I).c_str (), "r", stdin);
    freopen ((Name + _O).c_str (), "w", stdout);
    f[1] = 1, f[2] = 1;
    LL A, B, N; std :: cin >> A >> B >> N; rg int i;
    for (i = 3; i <= N; ++ i)
    {
        f[i] = (A * f[i - 1] + B * f[i - 2]) % 7;
        if (f[i] == 1 && f[i - 1] == 1) break;
    }
    if (i == 3) return puts ("1"), 0;
    std :: cout << ((i == N + 1) ? f[N] : f[((N % (i - 2)) == 0 ? (i - 2) : (N % (i - 2)))]);
    return 0;
}

 

/*
    矩阵快速幂 
*/
#include <cstdio>
#include <iostream>

#define rg register
typedef long long LL;

struct Matrix
{
    int c[2][2];
#define L 2
    Matrix operator * (const Matrix &rhs)
    {
        Matrix res;
        for (rg int i = 0, j, k; i < L; ++ i)
            for (j = 0; j < L; ++ j)
            {
                res.c[i][j] = 0;
                for (k = 0; k < L; ++ k)
                    (res.c[i][j] += c[i][k] * rhs.c[k][j]) %= 7;
            }
        res.c[0][0] %= 7;
        return res;
    }
} Answer, P;

std :: string Name = "attack", _I = ".in", _O = ".out";
int main (int argc, char *argv[])
{
    freopen ((Name + _I).c_str (), "r", stdin);
    freopen ((Name + _O).c_str (), "w", stdout);
    LL A, B, N; std :: cin >> A >> B >> N; rg int i;
    Answer.c[0][0] = Answer.c[0][1] = 1;
    Answer.c[1][1] = Answer.c[1][0] = 0;
    P.c[0][0] = A % 7, P.c[1][0] = B % 7, P.c[0][1] = 1, P.c[1][1] = 0;
    if (N <= 2) return puts ("1"), 0;
    for (N -= 2; N; P = P * P, N >>= 1)
        if (N & 1) Answer = Answer * P;
    std :: cout << Answer.c[0][0] % 7;
    return 0;
}

 

T2

 

/*
    求出公差即可 
*/
#include <cstdio>
#include <iostream>

#define rg register
inline void read (int &n)
{
    rg char c = getchar ();
    for (n = 0; !isdigit (c); c = getchar ());
    for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
}
typedef double flo;
std :: string Name = "fseq", _I = ".in", _O = ".out";
int main (int argc, char *argv[])
{
    freopen ((Name + _I).c_str (), "r", stdin);
    freopen ((Name + _O).c_str (), "w", stdout);
    int T; flo x, y, d; read (T);
    for (; T; -- T)
    {
        scanf ("%lf%lf", &x, &y);
        if (y == 0) { puts ("1.000000"); continue; }
        if (x == 0) { puts ("0.000000"); continue; }

        if (y > x) { puts ("0.000000"); continue; }
        d = 1.0 / (x + 1.0);
        printf ("%.6lf\n", 1.0 - y * d);
    }
    return 0;
}

 

 

T3

 

/*
    套路题，没营养
    Tarjan缩点 
    求出缩完点后树的直径
    对于每个点，若在最长链上就直接对两个端点取max
    否则就一直往上跳，累加边权，直到跳到最长链上
    然后做上一种情况的操作就好了 
*/  
#include <cstdio>
#include <iostream>
#define rg register

inline void read (int &n)
{
    rg char c = getchar ();
    for (n = 0; !isdigit (c); c = getchar ());
    for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
}

std :: string Name = "prize", _I = ".in", _O = ".out";

#define Max 40008
int _v[Max * 20], _n[Max * 20], list[Max * 20], EC = 1, _d[Max * 20];

int N;
inline void In (int u, int v, int d)
{ _v[++ EC] = v, _n[EC] = list[u], list[u] = EC, _d[EC] = d; }
int t, sk[Max], DC, SC;
int dfn[Max], low[Max], scc[Max];
 
inline void cmin (int &a, int b) { if (b < a) a = b; }

void Dfs (int n, int l)
{
    sk[++ t] = n; dfn[n] = low[n] = ++ DC;
    for (rg int i = list[n], v; i; i = _n[i])
        if (i != l && i != (l ^ 1))
        {
            if (!dfn[v = _v[i]]) Dfs (v, i), cmin (low[n], low[v]);
            else if (!scc[v]) cmin (low[n], dfn[v]);
        }
    if (low[n] == dfn[n])
    {
        ++ SC;
        for (int r = n + 1; t && r != n; -- t) 
            r = sk[t], scc[r] = SC;
    }
}

int ev[Max * 20], en[Max * 20], el[Max * 10], ec, ed[Max * 20];

inline void _In (int u, int v, int d)
{ 
    ev[++ ec] = v, en[ec] = el[u], el[u] = ec, ed[ec] = d; 
    ev[++ ec] = u, en[ec] = el[v], el[v] = ec, ed[ec] = d;
}

bool Find (int x, int y)
{
    for (rg int i = el[x]; i; i = en[i])
        if (ev[i] == y) return true;
    return false;
}

void Tarjan ()
{
    rg int i;
    for (i = 1; i <= N; ++ i) if (!dfn[i]) Dfs (i, 0);
    for (int n = 1; n <= N; ++ n)
        for (i = list[n]; i; i = _n[i])
            if (scc[n] != scc[_v[i]])
                if (!Find (scc[n], scc[_v[i]]))
                {
                    _In (scc[n], scc[_v[i]], _d[i]);
                }
}
int Answer[Max];
int dis1[Max], dis2[Max], pos1, pos2, Max1, Max2, pre[Max];
void Gc1 (int n, int F)
{
    for (rg int i = el[n], v; i; i = en[i])
        if ((v = ev[i]) != F) 
            dis1[v] = dis1[n] + ed[i], Gc1 (v, n);
    if (dis1[n] > Max1) Max1 = dis1[n], pos1 = n;
}
int f[Max];
void Gc2 (int n, int F)
{
    f[n] = F;
    for (rg int i = el[n], v; i; i = en[i])
        if ((v = ev[i]) != F) dis2[v] = dis2[n] + ed[i], Gc2 (v, n), pre[v] = n;
    if (dis2[n] > Max2) Max2 = dis2[n], pos2 = n;
}
bool isc[Max];
inline int max (int a, int b) { return a > b ? a : b; }
void GetAnswer ()
{
    Gc1 (1, 0), Gc2 (pos1, 0); rg int i;
    for (i = pos2; i; i = pre[i]) 
        isc[i] = true;
    rg int n; int res = 0, j;
    for (i = 1; i <= SC; ++ i)
    {
        res = 0;
        for (n = i; !isc[n] && n; n = f[n])
        {
            for (j = el[n]; j; j = en[j])
                if (ev[j] == f[n]) { res += ed[j]; break; }
        }
        res += max (dis2[n], Max2 - dis2[n]);
        Answer[i] = res;
    }
    for (i = 1; i <= N; ++ i)
        printf ("%d\n", Answer[scc[i]]);
}
int main (int argc, char *argv[])
{
    freopen ((Name + _I).c_str (), "r", stdin);
    freopen ((Name + _O).c_str (), "w", stdout);
    int x, y, M, z; read (N), read (M);
    for (rg int i = 1; i <= M; ++ i)
        read (x), read (y), read (z), In (x, y, z), In (y, x, z);
    
    Tarjan ();    
    
    GetAnswer ();
    
    return 0;
}