2018年浙江中医药大学程序设计竞赛 Solution

Problem A. Jhadgre的C语言程序

签。

 #include <bits/stdc++.h>
 using namespace std;

 int main()
 {
     puts("Helle World!");
     return ;
 }  

Problem B. Wpremig的AH之战

Solved.

• 如果 $n >= m 直接赢$
• 否则 $易知必败局面是当前剩余数是(n + 1)的倍数$

 #include<bits/stdc++.h>

 using namespace std;

 int n, m;

 int main()
 {
     while(~scanf("%d %d", &n, &m))
     {
         if(n >= m)
         {
             for(int i = m; i <= n; ++i)
                 printf("%d%c", i, " \n"[i == n]);
         }
         else
         {
             if(m % (n + ) == ) puts("You are loser");
             else printf("%d\n", m % (n + ));
         }
     }
     return ;
 }

Problem D. Jhadgre的梯子

温暖的签。

 #include <bits/stdc++.h>
 using namespace std;

 #define ll long long
 #define N 100010
 int n;
 ll m, x, Max;

 int main()
 {
     while (scanf("%d%lld", &n, &m) != EOF)
     {
         Max = ;
         for (int i = ; i <= n; ++i)
         {
             scanf("%lld", &x);
             Max = max(Max, x);
         }
         printf("%lld\n", max(0ll, Max - m));
     }
     return ;
 }

Problem E. Jhadgre的合唱队形

Solved.

题意：

有$n个队列，两种操作$

• $1\; x\; y\; z \;\;将编号属于[x, y]的队列最后加入一个z$
• $2\; x\; y\; z \;\; 询问编号属于[x, y]的所有队列中的所有人中第z大的数是多少$

思路：

本来想分块维护三个BIT,但是发现空间炸了。

那找第$k大的数可以二分，多次询问，就整体二分$

对于每个询问二分答案，在时间线中，在你前面的 并且 值小于你二分的答案就加入线段树

查询的时候直接查，记得更新k，然后将询问分组即可

 #include <bits/stdc++.h>
 using namespace std;

 #define ll long long
 #define N 50010
 int n, m, Q;

 namespace SEG
 {
     ll sum[N << ], lazy[N << ];
     void pushdown(int id, int l, int mid, int r)
     {
         if (!lazy[id]) return;
         sum[id << ] += 1ll * (mid - l + ) * lazy[id];
         lazy[id << ] += lazy[id];
         sum[id <<  | ] += 1ll * (r - mid) * lazy[id];
         lazy[id <<  | ] += lazy[id];
         lazy[id] = ;
     }
     void update(int id, int l, int r, int ql, int qr, int v)
     {
         if (l >= ql && r <= qr)
         {
             sum[id] += 1ll * v * (r - l + );
             lazy[id] += v;
             return;
         }
         int mid = (l + r) >> ;
         pushdown(id, l, mid, r);
         if (ql <= mid) update(id << , l, mid, ql, qr, v);
         if (qr > mid) update(id <<  | , mid + , r, ql, qr, v);
         sum[id] = sum[id << ] + sum[id <<  | ];
     }
     ll query(int id, int l, int r, int ql, int qr)
     {
 //      printf("%d %d %d %d\n", l, r, ql, qr);
         if (l >= ql && r <= qr) return sum[id];
         int mid = (l + r) >> ;
         pushdown(id, l, mid, r);
         ll res = ;
         if (ql <= mid) res += query(id << , l, mid, ql, qr);
         if (qr > mid) res += query(id <<  | , mid + , r, ql, qr);
         return res;
     }
 }

 struct node
 {
     int op, l, r, z, id;
     node () {}
     node (int op, int l, int r, int z, int id) : op(op), l(l), r(r), z(z), id(id) {}
 }q[N << ], ql[N << ], qr[N << ];
 int ans[N];

 void CDQ(int L, int R, int l, int r)
 {
     if (L > R) return;
     if (l == r)
     {
         for (int i = L; i <= R; ++i) if (q[i].op == )
             ans[q[i].id] = l;
         return;
     }
     int mid = (l + r) >> ;
     int cnt_ql = , cnt_qr = ;
     for (int i = L; i <= R; ++i)
     {
         if (q[i].op == )
         {
             if (q[i].z <= mid)
             {
                 SEG::update(, , n, q[i].l, q[i].r, );
                 ql[++cnt_ql] = q[i];
             }
             else
                 qr[++cnt_qr] = q[i];
         }
         else
         {
             ll sze = SEG::query(, , n, q[i].l, q[i].r);
             if (sze < q[i].z)
             {
                 q[i].z -= sze;
                 qr[++cnt_qr] = q[i];
             }
             else
                 ql[++cnt_ql] = q[i];
         }
     }
     for (int i = ; i <= cnt_ql; ++i) if (ql[i].z <= mid) SEG::update(, , n, ql[i].l, ql[i].r, -);
     for (int i = ; i <= cnt_ql; ++i) q[L + i - ] = ql[i];
     for (int i = ; i <= cnt_qr; ++i) q[L + cnt_ql + i - ] = qr[i];
     CDQ(L, L + cnt_ql - , l, mid);
     CDQ(L + cnt_ql, R, mid + , r);
 }

 int main()
 {
     while (scanf("%d%d", &n, &Q) != EOF)
     {
         m = ;
         memset(ans, , sizeof ans);
         for (int i = , op, x, y, z; i <= Q; ++i)
         {
             scanf("%d%d%d%d", &op, &x, &y, &z);
             if (op == ) q[++m] = node(, x, y, -z, i);
             else q[++m] = node(, x, y, z, i);
         }
         CDQ(, m, -n, -);
         for (int i = ; i <= Q; ++i) if (ans[i])
             printf("%d\n", -ans[i]);
     }
     return ;
 }  

Problem F. Jhadgre的伤心地

演了一波，小dp写成最短路、搜索可还行

 #include<bits/stdc++.h>

 using namespace std;

 typedef long long ll;

 const int maxn = 1e3 + ;

 int n;
 ll a[maxn];
 ll dis[maxn];
 int vis[maxn];

 void BFS()
 {
     memset(vis, , sizeof vis);
     memset(dis, 0x3f, sizeof dis);
     vis[] = ;
     dis[] = ;
     queue<int>q;
     q.push();
     while(!q.empty())
     {
         int u = q.front();
         q.pop();
         if(!vis[min(n, u + )])
         {
             dis[min(n, u + )] = dis[u] + ;
             vis[min(n, u + )] = ;
             q.push(min(n, u + ));
         }

         if(!vis[min(n * 1ll, a[u])])
         {
             dis[min(n * 1ll, a[u])] = dis[u] + ;
             vis[min(n * 1ll, a[u])] = ;
             q.push(min(n * 1ll, a[u]));
         }
     }
 }

 int main()
 {
     while(~scanf("%d", &n))
     {
         n++;
         for(int i = ; i <= n - ; ++i)
         {
             scanf("%d", a + i);
             a[i] = min(a[i], n * 1ll);
         }
         BFS();
         printf("%lld\n", dis[n]);
     }
     return ;
 }

Problem H. Jhadgre的回家之路

Solved.

两次搜索，搜'L' -> 'W' 和 'Q' -> 'W' 枚举'W'位置更新答案

薛佬交了三发，Rejudge后第一发其实就过

 #include<bits/stdc++.h>

 using namespace std;

 const int INF = 0x3f3f3f3f;
 const int maxn = 2e3 + ;

 int n, m;
 int vis[maxn][maxn];
 int dis1[maxn][maxn];
 int dis2[maxn][maxn];
 char mp[maxn][maxn];
 int dir[][] = {, ,  , ,-,  ,, -};

 struct node{
     int x, y;
     node(){}
     node(int x,int y):x(x), y(y){};
 }Q,L;

 bool check(int x, int y)
 {
     if(x <  || x > n || y <  || y > m || mp[x][y] == '#' || vis[x][y]) return false;
     else return true;
 }

 vector<node>W;

 void BFS1()
 {
     memset(vis, , sizeof vis);
     memset(dis1, INF, sizeof dis1);
     vis[L.x][L.y] = ;
     dis1[L.x][L.y] = ;
     queue<node>q;
     q.push(L);
     while(!q.empty())
     {
         node st = q.front();
         q.pop();
         for(int i =  ; i< ; ++i)
         {
             node now;
             now.x = st.x + dir[i][];
             now.y = st.y + dir[i][];
             if(check(now.x, now.y))
             {
                 vis[now.x][now.y] = ;
                 dis1[now.x][now.y] = dis1[st.x][st.y] + ;
                 q.push(now);
             }
         }
     }
 }

 void BFS2()
 {
     memset(vis, , sizeof vis);
     memset(dis2, INF, sizeof dis2);
     vis[L.x][L.y] = ;
     dis2[Q.x][Q.y] = ;
     queue<node>q;
     q.push(Q);
     while(!q.empty())
     {
         node st = q.front();
         q.pop();
         for(int i =  ; i< ; ++i)
         {
             node now;
             now.x = st.x + dir[i][];
             now.y = st.y + dir[i][];
             if(check(now.x, now.y))
             {
                 vis[now.x][now.y] = ;
                 dis2[now.x][now.y] = dis2[st.x][st.y] + ;
                 q.push(now);
             }
         }
     }
 }

 int main()
 {
     while(~scanf("%d %d", &n, &m))
     {
         for(int i = ; i <= n; ++i)
         {
             for(int j = ; j <= m; ++j)
             {
                 scanf(" %c", &mp[i][j]);
                 if(mp[i][j] == 'Q')
                 {
                     Q = node(i, j);
                 }
                 else if(mp[i][j] == 'L')
                 {
                     L = node(i, j);
                 }
                 else if(mp[i][j] == 'W')
                 {
                     W.push_back(node(i, j));
                 }
             }
         }
         BFS1();
         BFS2();
         int ans = INF;
         for(auto it: W)
         {
             ans = min(dis1[it.x][it.y] + dis2[it.x][it.y], ans);
         }
         printf("%d\n", ans);
     }
     return ;
 }

Problem I. Jhadgre的小饼干

string.find()

 #include<bits/stdc++.h>

 using namespace std;

 int n;
 string s;

 int main()
 {
     ios::sync_with_stdio(false);
     cin.tie();
     cout.tie();
     while(cin >> n)
     {
         int sum  = ;
         for(int i = ; i <= n; ++i)
         {
             cin >> s;
             if(s.find("zailaiyihe") != s.npos) sum++;
         }
         printf("%d\n", sum);
     }
     return ;
 }

Problem J. Jhadgre爬楼梯

线性推

 #include <bits/stdc++.h>
 using namespace std;

 #define ll long long
 #define N 10010
 const ll MOD = (ll);
 int n;
 ll f[N], sum[];

 int main()
 {
     f[] = ;
     sum[] = , sum[] = ;
     for (int i = ; i <= ; ++i)
     {
         f[i] = f[i - ];
         f[i] = (f[i] + sum[i & ]) % MOD;
         sum[i & ] = (sum[i & ] + f[i]) % MOD;
     }
     while (scanf("%d", &n) != EOF) printf("%lld\n", f[n]);
     return ;
 }

Problem L. Wpremig's Niuniu

题意：

规则：

五张牌均严格小于5，五张牌的和小于等于10，得60分。

五张牌都是$J,Q,K$，得50分。

五张牌有四张牌相同，得40分。

如果其中三张牌的和是10的倍数，并且其他两张牌的和是10的倍数，得30分

如果其中三张牌的和是10的倍数，并且其他两张牌的和对于10的余数大于等于7，得$2*$其他两张牌的和对于10的余数

如果其中三张牌的和是10的倍数，并且其他两张牌的和对于10的余数大于等于7，得其他两张牌的和对于10的余数

其余情况不得分，同时每次只能获取最高分。

现在你手上已经有四张牌，求拿到第五张牌后的分数期望（四舍五入）

思路：

预处理出所有情况，最后遍历一次得到总分后除以13

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
const ll MOD = 1e9 + ;

int ans[][][][][];

void Init()
{
    for (int a = ; a <= ; ++a)
    {
        for (int b = ; b <= ; ++b)
        {
            for (int c = ; c <= ; ++c)
            {
                for (int d = ; d <= ; ++d)
                {
                    for (int e = ; e <= ; ++e)
                    {
                        if (a <  && b <  && c <  && d <  && e <  && a + b + c + d + e <= )
                        {
                            ans[a][b][c][d][e] = ;
                        }
                        else if (a >=  && b >=  && c >=  && d >=  && e >= )
                        {
                            ans[a][b][c][d][e] = ;
                        }
                        else if ((a == b && b == c && c == d) || (a == b && b == c && c == e) || (a == b && b == d && d == e) || (a == c && c == d && d == e) || (b == c && c == d && d == e))
                        {
                            ans[a][b][c][d][e] = ;
                        }
                        else
                        {
                            if ((min(, a) + min(, b) + min(, c)) %  ==  || (min(, a) + min(, b) + min(, d)) %  ==  || (min(, a) + min(, b) + min(, e)) %  ==  || (min(, a) + min(, c) + min(, d)) %  ==  || (min(, a) + min(, c) + min(, e)) %  ==  || (min(, a) + min(, d) + min(, e)) %  ==
                                || (min(, b) + min(, c) + min(, d)) %  ==  || (min(, b) + min(, c) + min(, e)) %  ==  || (min(, b) + min(, d) + min(, e)) %  ==  || (min(, b) + min(, d) + min(, e)) %  ==
                                || (min(, c) + min(, d) + min(, e)) %  == )
                            {
                                int sum = min(, a) + min(, b) + min(, c) + min(, d) + min(, e);
                                if (sum %  == ) ans[a][b][c][d][e] = ;
                                else if (sum %  >= ) ans[a][b][c][d][e] =  * (sum % );
                                else ans[a][b][c][d][e] = sum % ;
                            }
                        }
                    }
                }
            }
        }
    }
}

void RUN()
{
    Init();
    int t;
    scanf("%d", &t);
    while (t--)
    {
        int a, b, c, d;
        scanf("%d %d %d %d", &a, &b, &c, &d);
        int sum = ;
        for (int i = ; i <= ; ++i) sum += ans[a][b][c][d][i];
        sum = (sum / 13.0) + 0.5;
        printf("%d\n", sum);
    }
}

int main()
{
#ifdef LOCAL_JUDGE
    freopen("Text.txt", "r", stdin);
#endif // LOCAL_JUDGE

    RUN();

#ifdef LOCAL_JUDGE
    fclose(stdin);
#endif // LOCAL_JUDGE
    return ;
}

枚举最后一张牌，求总分，最后除以13

#include<bits/stdc++.h>

using namespace std;

int a[], b[];
int c[];

void RUN()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        for (int i = ; i <= ; ++i) scanf("%d", a + i);
        b[] = min(, a[]) + min(, a[]);
        b[] = min(, a[]) + min(, a[]);
        b[] = min(, a[]) + min(, a[]);
        b[] = min(, a[]) + min(, a[]);
        b[] = min(, a[]) + min(, a[]);
        b[] = min(, a[]) + min(, a[]);
        b[] = ;
        int tmp = ;
        for (int i = ; i <= ; ++i) tmp += min(, a[i]);
        for (int i = ; i <= ; ++i)
        {
            if ((tmp - min(, a[i])) %  == )
            {
                b[] = ;
                break;
            }
        }

        double ans = ;
        for (int i = ; i <= ; ++i)
        {
            a[] = i;
            //
            if (a[] <  && a[] <  && a[] <  && a[] <  && a[] <  && a[] + a[] + a[] + a[] + a[] <= )
            {
                ans += 60.0;
                continue;
            }

            //
            if (a[] >=  && a[] >=  && a[] >=  && a[] >=  && a[] >= )
            {
                ans += 50.0;
                continue;
            }

            //
            if ((a[] == a[] && a[] == a[] && a[] == a[]) || (a[] == a[] && a[] == a[] && a[] == a[]) || (a[] == a[] && a[] == a[] && a[] == a[]) || (a[] == a[] && a[] == a[] && a[] == a[]) || (a[] == a[] && a[] == a[] && a[] == a[]))
            {
                ans += 40.0;
                continue;
            }

            //
            int sum = ;
            for (int i = ; i <= ; ++i) sum += min(, a[i]);
            if (b[])
            {
                if (sum %  == ) ans += 30.0;
                else if (sum %  >= ) ans += 2.0 * (sum % );
                else ans += sum % ;
                continue;
            }
            //
            int res = -;
            for (int j = ; j <= ; ++j)
            {
                if ((b[j] + min(, a[])) %  == )
                {
                    res = sum % ;
                    break;
                }
            }
            if (res == -) continue;
            if (res == ) ans += 30.0;
            else if (res >= ) ans += 2.0 * res;
            else if (res != -) ans += res;
        }
        ans = (ans / 13.0) + 0.5;
        printf("%d\n", (int)ans);
    }

}

int main()
{
#ifdef LOCAL_JUDGE
    freopen("Text.txt", "r", stdin);
#endif // LOCAL_JUDGE

    RUN();

#ifdef LOCAL_JUDGE
    fclose(stdin);
#endif // LOCAL_JUDGE
    return ;
}

比赛的时候写丑了，同时对于题意理解不明确？（说好的一副牌呢！）