HNOI[2008]题解

最新推荐文章于 2023-01-13 15:48:19 发布

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刷完了HNOI[2008]，来写个题解吧。

Cards

#include <cstdio>
#include <vector>
#include <cstring>
#define pb push_back
using namespace std;

const int MAXM = 69;
const int MAXC = 29;

int sr;
int sg;
int sb;
int m;
int p;
int n;

vector<int> cir[MAXM];

int ans;
int f[MAXC][MAXC][MAXC];

inline int adv(int m, int p)
{
    int re = 1, base = m;
    int tp = p-2;
    while(tp)
    {
        if(tp&1) re = re*base%p;
        tp >>= 1;
        base = base*base%p; 
    }
    return re;
}

int main()
{
    scanf("%d%d%d%d%d", &sr, &sg, &sb, &m, &p);
    n = sr+sg+sb;
    for(int i = 1; i <= m; ++i)
    {
        int g[MAXM] = {0};
        bool vis[MAXM] = {false};
        for(int j = 1; j <= n; ++j)
            scanf("%d", g+j);
        for(int j = 1; j <= n; ++j)
            if(!vis[j])
            {
                int sum = 0;
                int k = j;
                while(!vis[k])
                {
                    sum++;
                    vis[k] = true;
                    k = g[k];
                }
                cir[i].pb(sum);
            }
    }
    m++;
    for(int i = 1; i <= n; ++i)
        cir[m].pb(1);
    for(int i = 1; i <= m; ++i)
    {
        memset(f, 0, sizeof(f));
        f[0][0][0] = 1;
        for(int j = 1, sz = cir[i].size(); j <= sz; ++j)
        {
            int w = cir[i][j-1];
            for(int r = sr; r >= 0; --r)
                for(int g = sg; g >= 0; --g)
                    for(int b = sb; b >= 0; --b)
                    {
                        if(r >= w) f[r][g][b] += f[r-w][g][b];
                        if(g >= w) f[r][g][b] += f[r][g-w][b];
                        if(b >= w) f[r][g][b] += f[r][g][b-w];
                        f[r][g][b] %= p;
                    }
        }
        ans = (ans+f[sr][sg][sb])%p;
    }
    ans = ans*adv(m, p)%p;
    printf("%d", ans);
    return 0;
}

明明的烦恼

题解请走这里：
http://www.cnblogs.com/zhj5chengfeng/p/3278557.html

#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 1009;
const int MAXL = 1009;

int n;
int d[MAXN];
int sum;
int cnt;

struct bi
{
    int len;
    int a[MAXL];
    inline void reset(int k)
    {
        len = 0;
        memset(a, 0, sizeof(a));
        a[++len] = k;   
    }
    inline void operator *= (int k)
    {
        for(int i = 1; i <= len; ++i)
            a[i] *= k;
        for(int i = 1; i <= len; ++i)
        {
            a[i+1] += a[i]/10000;
            a[i] %= 10000;
        }
        while(a[len+1])
        {
            len++;
            a[len+1] += a[len]/10000;
            a[len] %= 10000;
        }
    }
    inline void operator /= (int k)
    {
        for(int i = len; i >= 1; --i)
        {
            a[i-1] += a[i]%k*10000; 
            a[i] /= k;
        }
        while(!a[len])
            len--;
    }
    inline void print()
    {
        printf("%d", a[len]);
        for(int i = len-1; i >= 1; --i)
            printf("%04d", a[i]);   
    }
}ans;

int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
    {
        scanf("%d", d+i);
        if(d[i] != -1)
        {
            cnt++;
            sum += d[i]-1;  
        }
    }
    ans.reset(1);
    for(int i = n-2; i > n-2-sum; --i)
        ans *= i;
    for(int i = 1; i <= n; ++i)
        if(d[i] != -1)
            for(int j = 2; j < d[i]; ++j)
                ans /= j;
    for(int i = 1; i <= n-2-sum; ++i)
        ans *= (n-cnt);
    ans.print();
    puts("");
    return 0;
}

神奇的国度

基础知识请走这里：
http://blog.youkuaiyun.com/qq_20118433/article/details/44420099
完美消除序列从后往前依次给每个点染色，给每个点染上可以染的最小的颜色。
证明：
设使用 $T$ 种颜色, 则 $T ≥$ 色数
$T =$ 团数 $≤$ 色数
团数 = 色数 = $T$
时间复杂度： $O(m + n)$

#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
const int MAXN = 10009, MAXM = 1000009;

int n, m;

inline int readi()
{
    int re = 0;
    bool flag = false;
    char c = getchar();
    while(1)
    {
        if(c >= '0' && c <= '9')
        {
            re = re*10+c-'0';
            flag = true;
        }
        else 
            if(flag) return re;
        c = getchar();
    }
}

struct Graph
{
    int p;
    int head[MAXN];
    int v[MAXM<<1];
    int ne[MAXM<<1];
    Graph(){p = 1;}
    inline void add(int u, int vv)
    {
        v[p] = vv;ne[p] = head[u];
        head[u] = p++;  
    }
}E;

int seg[MAXN];
bool vis[MAXN];
int col[MAXN];
int hash[MAXN];
int Max = 1;

int ans;

struct List
{
    int id;
    List *ne, *fr;  
}*label[MAXN] = {NULL}, *tail[MAXN] = {NULL}, null = {0, NULL, NULL};

int line[MAXN];
List *place[MAXN];

inline void ins(int i, int p)
{
    List *pre = tail[i]->fr, *bac = tail[i], *now;
    now = (List *)malloc(sizeof(List));
    *now = null;now->id = p;
    pre->ne = now;now->fr = pre;
    bac->fr = now;now->ne = bac;
    place[p] = now;
    line[p] = i;
}

inline void del(List *now)
{
    List *pre = now->fr, *bac = now->ne;
    pre->ne = bac;bac->fr = pre;
    line[now->id] = 0;
    place[now->id] = NULL;
    free(now);
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; ++i)
    {
        int u, v;
        u = readi();
        v = readi();
        E.add(u, v);E.add(v, u);    
    }

    for(int i = 1; i <= n; ++i)
    {
        label[i] = (List *)malloc(sizeof(List));
        *label[i] = null;
        label[i]->ne = (List *)malloc(sizeof(List));
        *label[i]->ne = null;
        tail[i] = label[i]->ne;
        label[i]->ne = tail[i];
        tail[i]->fr = label[i];
        ins(1, i);
    }

    for(int i = n; i >= 1; --i)
    {
        List *p;
        for(List *j = label[Max]->ne; ;)
        {
            while(j == tail[Max]) j = label[--Max]->ne;
            if(vis[j->id])
            {
                List *tmp = j->ne;
                del(j);
                j = tmp;
            }
            else 
            {
                p = j;
                break;
            }
        }
        seg[i] = p->id;
        vis[p->id] = true;
        for(int j = E.head[p->id]; j; j = E.ne[j])
            if(!vis[E.v[j]])
            {
                int row = line[E.v[j]];
                ins(row+1, E.v[j]);
                if(Max < row+1) Max = row+1;
            }
        del(p);
    }

    memset(vis+1, false, n*sizeof(bool));
    for(int i = n; i >= 1; --i)
    {
        vis[seg[i]] = true;
        col[seg[i]] = 1;
        for(int j = E.head[seg[i]]; j; j = E.ne[j])
            if(vis[E.v[j]])
                hash[col[E.v[j]]] = i;
        for(int j = 1; j <= n; ++j)
            if(hash[j] != i)
            {
                col[seg[i]] = j;
                break;  
            }
        if(col[seg[i]] > ans) ans = col[seg[i]];
    }

    printf("%d\n", ans);
    return 0;
}

水平可见直线

这里写图片描述

#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define sqr(x) ((x)*(x))
using namespace std;
const int maxn = 50009;
const double eps = 1e-8;
int n;
struct Line
{
    int id;
    double k, b;
}line[maxn];
int top;
Line stack[maxn];
bool visable[maxn];

inline bool cmpk(const Line &a, const Line &b)
{
    if(fabs(a.k-b.k) <= eps) return a.b < b.b;
    else return a.k < b.k;
}

inline double interx(const Line &a, const Line &b)
{
    return (b.b-a.b)/(a.k-b.k); 
}

inline void insert(const Line &a)
{
    while(top)
    {
        Line t = stack[top], t_1 = stack[top-1];
        if(fabs(a.k-t.k) <= eps) --top;
        else if(top > 1 && interx(a, t) <= interx(t, t_1)) --top;
        else break;
    }
    stack[++top] = a;
}

int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
    {   
        scanf("%lf%lf", &line[i].k, &line[i].b);
        line[i].id = i;
    }
    sort(line+1, line+n+1, cmpk);
    for(int i = 1; i <= n; ++i)
        insert(line[i]);
    for(int i = 1; i <= top; ++i)
        visable[stack[i].id] = true;
    for(int i = 1; i <= n; ++i)
        if(visable[i]) printf("%d ", i);
    return 0;   
}

越狱

这里写图片描述

#include <iostream>
const int mod = 100003;

inline int power(int a, long long b)
{
    int re = 1, base = a;
    while(b)
    {
        if(b&1) re = (long long)re*base%mod;
        base = (long long)base*base%mod;
        b >>= 1;
    }
    return re;
}

int main()
{
    int m = 0;
    long long n = 0;
    std::cin >> m >> n;
    m %= mod;
    std::cout << (power(m, n)-(long long)m*power(m-1, n-1)%mod+mod)%mod << std::endl;
    return 0;   
}

GT考试

这里写图片描述

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
using namespace std;

const int maxm = 29;

int n, m, mod;
string a;
int f[maxm];

struct mtx
{
    int g[maxm][maxm];
    mtx(){memset(g, 0, sizeof(g));}
    inline mtx operator * (const mtx &a)
    {
        mtx b;
        for(int i = 0; i < m; ++i)
            for(int j = 0; j < m; ++j)
                for(int k = 0; k < m; ++k)
                    b.g[i][j] = (b.g[i][j]+g[i][k]*a.g[k][j]%mod)%mod;
        return b;
    }
}M, ans;

int main()
{
    cin >> n >> m >> mod >> a;
    for(int i = 1; i < m; ++i)
    {
        int j = f[i];
        while(j && a[j] != a[i]) j = f[j];
        f[i+1] = a[j]==a[i] ? j+1 : 0;
    }
    for(int i = 0; i < m; ++i)
        for(int j = '0'; j <= '9'; ++j)
        {
            if((char)j == a[i]) M.g[i][i+1]++;
            else
            {
                int k = i;
                while(k && a[k] != (char)j) k = f[k];
                if(a[k] == (char)j) k++;
                M.g[i][k]++;
            }
        }
    for(int i = 0; i < m; ++i)
        ans.g[i][i] = 1;
    int b = n;
    while(b)
    {
        if(b&1) ans = ans*M;
        M = M*M;
        b >>= 1;
    }
    int sum = 0;
    for(int i = 0; i < m; ++i)
        sum = (sum+ans.g[0][i])%mod;
    cout << sum << endl;    
    return 0;
}

玩具装箱toy

这里写图片描述

#include <cstdio>
#define sqr(x) ((x)*(x))
using namespace std;
typedef long long LL;
const int maxn = 50009;
int n, L;
LL c[maxn];
LL sum[maxn];
LL f[maxn];
int q[maxn];
int l, r;

#define T(p) (sum[p]+p)
#define X(p) (T(p)<<1)
#define Y(p) (f[p]+sqr(T(p)+L+1))

int main()
{
    scanf("%d%d", &n, &L);
    for(int i = 1; i <= n; ++i)
    {
        scanf("%lld", c+i);
        sum[i] = sum[i-1]+c[i];
    }
    for(int i = 1; i <= n; ++i)
    {
        while(l <= r-1 && Y(q[l+1])-Y(q[l]) <= T(i)*(X(q[l+1])-X(q[l])))
            q[l++] = 0;
        f[i] = f[q[l]]+sqr(i-q[l]-1+sum[i]-sum[q[l]]-L);
        while(l <= r && (Y(i)-Y(q[r]))*(X(i)-X(q[r-1])) < (Y(i)-Y(q[r-1]))*(X(i)-X(q[r])))
            q[r--] = 0;
        q[++r] = i;
    }
    printf("%lld\n", f[n]);
    return 0;
}

遥远的行星

这里写图片描述

#include <cstdio>
#include <cmath>
using namespace std;
const int maxn = 100009;
const double eps = 1e-6;
int n;
double a;
double m[maxn];
double sum[maxn];

int main()
{
    scanf("%d%lf", &n, &a);
    for(int i = 1; i <= n; ++i)
    {
        scanf("%lf", m+i);
        sum[i] = sum[i-1]+m[i];
    }
    for(int i = 1; i <= n; ++i)
    {
        int k = a*i;
        double f = 0;
        if(fabs((k+1)*1.0-a*i) <= eps) k++;
        if(k <= 50)
            for(int j = 1; j <= k; ++j)
                f += m[i]*m[j]/(double)(i-j);
        else f = sum[k]*m[i]/(double)(i-((1+k)>>1));
        printf("%lf\n", f);
    }
    return 0;
}

差不多就是这样了，HNOI[2009]走起…