codeforces educational round 25

A 出题人不给样例解释。。。具体程序

#include<bits/stdc++.h>
using namespace std;
int n;
char s[];
int main()
{
    scanf("%d%s", &n, s + );
    int ans = , tot = ;
    for(int i = ; i <= n; ++i)
    {
        if(s[i] == '') ++tot;
        else
        {
            ans = ans *  + tot;
            tot = ;
        }
    }
    printf("%d\n", ans *  + tot);
    return ;
}

B 枚举每个点填上黑子，然后判断

#include<bits/stdc++.h>
using namespace std;
const int dx[] = {, , -, -}, dy[] = {, , , };
int n;
char Map[][];
bool check()
{
    for(int i = ; i <= ; ++i)
        for(int j = ; j <= ; ++j) if(Map[i][j] == 'X')
        {
            for(int k = ; k < ; ++k)
            {
                int xx = i, yy = j;
                bool flag = true;
                for(int l = ; l < ; ++l)
                {
                    xx += dx[k], yy += dy[k];
                    if(Map[xx][yy] != 'X')
                    {
                        flag = false;
                        break;
                    }
                }
                if(flag) return true;
            }
        }
    return false;
}
int main()
{
    memset(Map, 'O', sizeof(Map));
    for(int i = ; i <= ; ++i)
        scanf("%s", Map[i] + );
    for(int i = ; i <= ; ++i)
        for(int j = ; j <= ; ++j) if(Map[i][j] == '.')
        {
            char c = Map[i][j];
            Map[i][j] = 'X';
            if(check())
            {
                puts("YES");
                return ;
            }
            Map[i][j] = c;
        }
    puts("NO");
    return ;
}

C 贪心，不够就乘2，每次和a取大

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = ;
int n, ans;
ll k;
ll a[N];
int main()
{
    scanf("%d%d", &n, &k);
    for(int i = ; i <= n; ++i) scanf("%lld", &a[i]);
    sort(a + , a + n + );
    for(int i = ; i <= n; ++i)
    {
        while(k < (a[i] + 1ll) / 2ll)
        {
            ++ans;
            k = 2ll * k;
        }
        if(k >= (a[i] + 1ll) / 2ll) k = max(k, a[i]);
    }
    printf("%d\n", ans);
    return ;
}

D 贪心，能互相交换肯定能换成那个目标串，所以看现在缺少目标串哪个元素就加上

#include<bits/stdc++.h>
using namespace std;
const int N = ;
char s[N], t[N];
int cnt[], cnt1[];
int main()
{
    scanf("%s%s", s + , t + );
    int len1 = strlen(t + ), len2 = strlen(s + );
    for(int i = ; i <= len1; ++i) ++cnt[t[i] - 'a'];
    for(int i = ; i <= len2; ++i) if(s[i] != '?')
        ++cnt1[s[i] - 'a'];
    for(int i = ; i <= len2; ++i)
    {
        bool flag = false;
        if(s[i] == '?')
        {
            int pos = ;
            for(int j = ; j < ; ++j) if(cnt1[j] < cnt[j])
            {
                ++cnt1[j];
                flag = true;
                pos = j;
                break;
            }
            if(!flag)
            {
                for(int j = ; j < ; ++j) cnt1[j] -= cnt[j];
                for(int j = ; j < ; ++j) if(cnt1[j] < cnt[j])
                {
                    ++cnt1[j];
                    pos = j;
                    break;
                }
            }
            printf("%c", (char)(pos + 'a'));
        }
        else printf("%c", s[i]);
    }
    return ;
}

E 贪心，证明待填

#include<bits/stdc++.h>
using namespace std;
const int N = ;
int n, m;
priority_queue<int> q;
vector<int> G[N];
int c[N], d[N];
int main()
{
    scanf("%d%d", &n, &m);
    for(int i = ; i <= m; ++i)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        G[v].push_back(u);
        ++d[u];
    }
    for(int i = ; i <= n; ++i) if(!d[i]) q.push(i);
    for(int i = n; i; --i)
    {
        int x = q.top();
        q.pop();
        c[x] = i;
        for(int j = ; j < G[x].size(); ++j)
        {
            int v = G[x][j];
            --d[v];
            if(!d[v]) q.push(v);
        }
    }
    for(int i = ; i <= n; ++i) printf("%d ", c[i]);
    return ;
}

F dp dp[i]=min(dp[j-1]+calc(j, i))，calc可以先预处理lcp，然后暴力枚举串的长度，看是否符合 这个跑不过去，要4s

#include<bits/stdc++.h>
using namespace std;
const int N = ;
int n;
int dp[N][N], f[N], sum[N];
char s[N];
int getlen(int x)
{
    int ret = ;
    while(x)
    {
        ++ret;
        x /= ;
    }
    return ret;
}
int main()
{
    scanf("%s", s + );
    n = strlen(s + );
    for(int i = n; i; --i)
        for(int j = n; j >= i; --j)
            if(s[i] == s[j])
                dp[i][j] = dp[i + ][j + ] + ;
    memset(f, 0x3f3f, sizeof(f));
    f[] = ;
    for(int i = ; i <= ; ++i) sum[i] = sum[i / ] + ;
    for(int i = ; i <= n; ++i)
        for(int j = i; j; --j)
            for(int k = i - j + ; k >  && dp[k][i - j + ] >= j; k -= j)
                f[i] = min(f[i], f[k - ] + j + sum[(i - k + ) / j]);
    printf("%d\n", f[n]);
    return ;
}

G dfs 思路很好，把第一个变成黑点的点作为根，然后预处理出每个点到根的最小值，每次查询直接拿自己的最小值和上次答案比较，因为上次的答案这次肯定也能用，更新就是把当前的最小值和ans取min

#include<bits/stdc++.h>
using namespace std;
const int N = ;
int opt, x, n, q, last, ans =  << ;
int up[N];
vector<int> G[N];
void dfs(int u, int last)
{
    for(int i = ; i < G[u].size(); ++i)
    {
        int v = G[u][i];
        if(v == last) continue;
        up[v] = min(v, up[u]);
        dfs(v, u);
    }
}
int main()
{
    scanf("%d%d", &n, &q);
    for(int i = ; i < n; ++i)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    scanf("%d%d", &opt, &x);
    x = (x + last) % n + ;
    up[x] = x;
    dfs(x, );
    q--;
    while(q--)
    {
        scanf("%d%d", &opt, &x);
        x = (x + last) % n + ;
        if(opt == ) ans = min(ans, up[x]);
        if(opt == )
        {
            printf("%d\n", min(ans, up[x]));
            last = min(ans, up[x]);
        }
    }
    return ;
}