第十八周 12.27-1.2

因为期末了本来不想写了 但是觉得挂篇空文也无所谓 还是写了

12.27

去西工大玩。

12.28-12.29

什么都没干。

12.30

上次挂的BC。

HDU 5602 Black Jack

一开始以为A用1表示。

WA了pretest。

看clar改完过了pre然后fst。

一直以为是dp错。找了很久很久找不出来。

后来看div2的clar才知道只有10用T表示。JQK还是JQK。

这样都能过pre。sample也没有。无语了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
double dp1[22][22], dp2[22][22];//x z
double p[11] = { 0,
    1.0 / 13, 1.0 / 13, 1.0 / 13,
    1.0 / 13, 1.0 / 13, 1.0 / 13,
    1.0 / 13, 1.0 / 13, 1.0 / 13, 4.0 / 13
};

void pre()
{
    for(int i = 2; i <= 21; i++)
        for(int j = i; j <= 21; j++)
            dp1[i][j] = 1;

    for(int i = 21; i >= 2; i--)
        for(int j = i - 1; j >= 2; j--)
            for(int k = 10; k >= 1; k--)
                if(j + k <= 21) dp1[i][j] += p[k] * dp1[i][j+k];

    for(int i = 21; i >= 2; i--)
    {
        for(int j = 21; j >= 2; j--)
        {
            double tmp = 0;//draw->lose
            for(int k = 10; k >= 1; k--)
            {
                if(i + k > 21) tmp += p[k];
                else tmp += p[k] * dp2[i+k][j];
            }
            dp2[i][j] = min(dp1[i][j], tmp);
        }
    }
    return;
}

int num(char c)
{
    if(c == 'A') return 1;
    if(c >= '0' && c <= '9') return c - '0';
    return 10;
}

int main(void)
{
    pre();
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int s1, s2;
        char s[11];
        scanf("%s", s);
        s1 = num(s[0]) + num(s[1]);
        s2 = num(s[2]) + num(s[3]);
        puts(dp2[s1][s2] < 0.5 ? "YES" : "NO");
    }
    return 0;
}

12.31

CF 611 D New Year and Ancient Prophecy

n2搞下LCP就好了。QAQ

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = 1e9 + 7;
LL dp[5005][5005];
char num[5005];
int com[5005][5005];

bool bigger(int pos1, int pos2, int len)
{
    int x = com[pos1][pos2];
    if( x >= len) return false;
    return num[pos1+x] > num[pos2+x];
}

int main(void)
{
    int n;
    scanf("%d%s", &n, num + 1);

    for(int i = n; i >= 1; i--)
        for(int j = n; j >= 1; j--)
            com[i][j] = num[i] == num[j] ? ( com[i+1][j+1] + 1 ) : 0 ;

    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= i; j++)
        {
            if(num[i-j+1] == '0') dp[i][j] = dp[i][j-1];
            else if(i == j) dp[i][j] = ( dp[i][j-1] + 1LL ) % mod;
            else if( i >= 2 * j && bigger(i-j+1, i-2*j+1, j) ) dp[i][j] = ( dp[i][j-1] + dp[i-j][j] ) % mod;
            else dp[i][j] = ( dp[i][j-1] + dp[i-j][min(i-j,j-1)]) % mod;
        }
    }
    printf("%I64d
", dp[n][n]);
    return 0;
}

1.1

CF 611 E New Year and Three Musketeers

多重集搞搞。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <set>
using namespace std;
multiset<int> M;
multiset<int>::iterator it;
int m[3];

int main(void)
{
    int n;
    scanf("%d", &n);
    for(int i = 0; i < 3; i++) scanf("%d", m + i);
    sort(m, m + 3);
    int a = m[0], b = m[1], c = m[2];

    for(int i = 0; i < n; i++)
    {
        int t;
        scanf("%d", &t);
        if(t > a + b + c) {puts("-1"); return 0;}
        M.insert(t);
    }

    int ans = 0;

    while(!M.empty())
    {
        it = M.lower_bound(b + c + 1);
        if(it == M.end()) break;
        M.erase(it); ans++;
    }

    while(!M.empty())
    {
        it = M.lower_bound(a + c + 1);
        if(it == M.end()) break;
        M.erase(it); ans++;
        if(M.empty()) break;
        it = M.lower_bound(a + 1);
        if(it != M.begin()) M.erase(--it);
    }

    while(!M.empty())
    {
        it = M.lower_bound(max(c, a + b) + 1);
        if(it == M.end()) break;
        M.erase(it); ans++;
        if(M.empty()) break;
        it = M.lower_bound(b + 1);
        if(it != M.begin()) M.erase(--it);
    }

    int x = 0, y = 0, k = 0;
    for(it = M.begin(); it != M.end(); it++)
    {
        if(*it <= a + b) x++;
        if(*it <= c) y++;
    }

    int s = max( (max(x, y) + 1) /2, max(x-y, y-x) );
    while(!M.empty() && y)
    {
        it = M.lower_bound(c + 1);
        if(it != M.begin())
        {
            --it;
            if(*it <= a + b) x--;
            if(*it <= c) y--;
            M.erase(it);
        }
        it = M.lower_bound(b + 1);
        if(it != M.begin())
        {
            --it;
            if(*it <= a + b) x--;
            if(*it <= c) y--;
            M.erase(it);
        }
        it = M.lower_bound(a + 1);
        if(it != M.begin())
        {
            --it;
            if(*it <= a + b) x--;
            if(*it <= c) y--;
            M.erase(it);
        }
        s = min( s, ++k + max( (max(x, y) + 1)/2, max(x-y, y-x) ) );
    }

    printf("%d
", ans + s);

    return 0;
}

1.2

矩阵快速幂！矩阵快速幂！矩阵快速幂！

一直懒得自己码板。每次都是用到去搜。受不了了！

struct Matrix
{
    LL m[maxn][maxn];
    Matrix(){memset(m, 0, sizeof(m));}
    void E(){for(int i = 0; i < maxn; i++) m[i][i] = 1;}
};

Matrix M_mul(Matrix a, Matrix b)
{
    Matrix ret;
    for(int i = 0; i < maxn; i++)
        for(int j = 0; j < maxn; j++)
            for(int k = 0; k < maxn; k++)
                ret.m[i][j] = ( ret.m[i][j] + (a.m[i][k] * b.m[k][j]) % mod ) % mod;
    return ret;
}

Matrix M_qpow(Matrix P, LL n)
{
    Matrix ret;
    ret.E();
    while(n)
    {
        if(n & 1LL) ret = M_mul(ret, P);
        n >>= 1LL;
        P = M_mul(P, P);
    }
    return ret;
}

HDU 5607 graph

直接搞！

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 55;
const LL mod = 1e9 + 7;
int G[maxn][maxn], deg[maxn];
LL ans[22][maxn];

struct query
{
    int id, u;
    LL K;
}q[22];

bool cmp(query a, query b)
{
    return a.K < b.K;
}

LL qpow(LL a, LL b)
{
    LL ret = 1LL, tmp = a;
    while(b)
    {
       if(b & 1LL) ret = ret * tmp % mod;
       tmp = tmp * tmp % mod;
       b >>= 1LL;
    }
    return ret;
}

LL inv(LL a)
{
    LL k = 1e9 + 5;
    return qpow(a, k);
}

struct Matrix
{
    LL m[maxn][maxn];
    Matrix(){memset(m, 0, sizeof(m));}
    void E(){for(int i = 0; i < maxn; i++) m[i][i] = 1;}
};

Matrix M_mul(Matrix a, Matrix b)
{
    Matrix ret;
    for(int i = 0; i < maxn; i++)
        for(int j = 0; j < maxn; j++)
            for(int k = 0; k < maxn; k++)
                ret.m[i][j] = ( ret.m[i][j] + (a.m[i][k] * b.m[k][j]) % mod ) % mod;
    return ret;
}

Matrix M_qpow(Matrix P, LL n)
{
    Matrix ret;
    ret.E();
    while(n)
    {
        if(n & 1LL) ret = M_mul(ret, P);
        n >>= 1LL;
        P = M_mul(P, P);
    }
    return ret;
}

int main(void)
{
    int N, M;
    scanf("%d%d", &N, &M);
    for(int i = 1; i <= M; i++)
    {
        int X, Y;
        scanf("%d%d", &X, &Y);
        deg[X]++;
        G[X][Y] = 1;
    }
    int Q;
    scanf("%d", &Q);
    for(int i = 0; i < Q; i++)
    {
        scanf("%d%I64d", &q[i].u, &q[i].K);
        q[i].id = i;
    }
    sort(q, q + Q, cmp);
    Matrix A, B = M_qpow(A, 0LL);
    for(int i = 1; i <= N; i++)
    {
        LL R = inv(deg[i]);
        for(int j = 1; j <= N; j++)
            if(G[i][j]) A.m[j][i] = R;
    }
    for(int i = 0; i < Q; i++)
    {
        LL nn;
        if(!i) nn = q[i].K;
        else nn = q[i].K - q[i-1].K;
        B = M_mul(B, M_qpow(A, nn));
        for(int j = 1; j <= N; j++) ans[q[i].id][j] = B.m[j][q[i].u];
    }
    for(int i = 0; i < Q; i++)
    {
        for(int j = 1; j <= N; j++) printf("%d ", ans[i][j]);
        puts("");
    }
    return 0;
}