Codeforces_828

A.模拟，注意单人的时候判断顺序。

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

int n,a,b;

int main()
{
    ios::sync_with_stdio();
    cin >> n >> a >> b;
    int b1 = b,b2 = ,ans = ;
    for(int i = ;i <= n;i++)
    {
        int x;
        cin >> x;
        if(x == )
        {
            if(a)    a--;
            else if(b1)
            {
                b1--;
                b2++;
            }
            else if(b2)    b2--;
            else    ans++;
        }
        else
        {
            if(b1)    b1--;
            else    ans += ;
        }
    }
    cout << ans << endl;
    return ;
}

---

B.对于每个B点，更新最极端边界，确定最后的长宽，注意不用涂和不成立的情况。

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

int n,m;
string s[];

int main()
{
    ios::sync_with_stdio();
    cin >> n >> m;
    for(int i = ;i <= n;i++)
    {
        cin >> s[i];
        s[i] = ' '+s[i];
    }
    int l = ,r = ,h = ,d = ,cnt = ;
    for(int i = ;i <= n;i++)
    {
        for(int j = ;j <= m;j++)
        {
            if(s[i][j] == 'W')    continue;
            cnt++;
            l = min(l,j);
            r = max(r,j);
            h = max(h,i);
            d = min(d,i);
        }
    }
    int ans = max(r-l+,h-d+);
    if(ans <= )
    {
        cout <<  << endl;
        return ;
    }
    if(ans > n || ans > m)
    {
        cout << - << endl;
        return ;
    }
    cout << ans*ans-cnt << endl;
    return ;
}

---

C.对于每一个串，选择更后面的起点更新。

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

char a[] = "";
int n;

int main()
{
    ios::sync_with_stdio();
    cin >> n;
    int len = ;
    while(n--)
    {
        string s;
        int x,now = ;
        cin >> s >> x;
        while(x--)
        {
            int xx;
            cin >> xx;
            int i = max(now,xx);
            for(int j = i-xx;j < s.length();i++,j++)
            {
                len = max(len,i);
                a[i] = s[j];
            }
            now = max(now,i);
        }
    }
    for(int i = ;i <= len;i++)
    {
        if(a[i])    cout << a[i];
        else    cout << 'a';
    }
    cout << endl;
    return ;
}

---

D.从一点个拉出k条链来，每条链长度尽可能相等。

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

int n,k;

int main()
{
    ios::sync_with_stdio();
    cin >> n >> k;
    int t = n-k-;
    if(t%k == )    cout << +t/k* << endl;
    else if(t%k == )   cout << +t/k*+ << endl;
    else    cout << +t/k*+ << endl;
    int i;
    for(i = n-;i >= max(n-k,);i--)    cout << n << " " << i << endl;
    for(;i >= ;i--)   cout << i+k << " " << i << endl;
    return ;
}

---

E.因为询问的串长最多为10，我们可以每个字母对应的长度都开个树状数组。

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

string s;
int q,tree[][][][] = {};
map<char,int> mp;

inline int lowbit(int x)
{
    return x&-x;
}

void add(int x,int y,int z,int pos,int xx)
{
    while(pos < s.length())
    {
        tree[x][y][z][pos] += xx;
        pos += lowbit(pos);
    }
}

int getsum(int x,int y,int z,int pos)
{
    int sum = ;
    while(pos > )
    {
        sum += tree[x][y][z][pos];
        pos -= lowbit(pos);
    }
    return sum;
}

int main()
{
    ios::sync_with_stdio();
    cin >> s >> q;
    s = ' '+s;
    mp['A'] = ;
    mp['G'] = ;
    mp['C'] = ;
    mp['T'] = ;
    for(int i = ;i < s.length();i++)
    {
        for(int j = ;j <= ;j++)  add(mp[s[i]],i%j,j,i,);
    }
    while(q--)
    {
        int x;
        cin >> x;
        if(x == )
        {
            string ss;
            cin >> x >> ss;
            for(int i = ;i <= ;i++)  add(mp[s[x]],x%i,i,x,-);
            for(int i = ;i <= ;i++)  add(mp[ss[]],x%i,i,x,);
            s[x] = ss[];
        }
        else
        {
            int l,r;
            string ss;
            cin >> l >> r >> ss;
            int sum = ;
            for(int i = ;i < ss.length()&&l+i <= r;i++)    sum += getsum(mp[ss[i]],(l+i)%ss.length(),ss.length(),r)-getsum(mp[ss[i]],(l+i)%ss.length(),ss.length(),l-);
            cout << sum << endl;
        }
    }
    return ;
}