Codeforces_794

A.统计两个guard之间的钞票数。

#include<bits/stdc++.h>
#define MOD 1000000009
using namespace std;

int a,b,c,n;

int main()
{
    ios::sync_with_stdio(false);
    cin >> a >> b >> c >> n;
    int ans = ;
    while(n--)
    {
        int x;
        cin >> x;
        if(b < x && x < c)  ans++;
    }
    cout << ans << endl;
    return ;
}

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B.面积x倍，边长sqrt(x)倍。

#include<bits/stdc++.h>
#define MOD 1000000009
using namespace std;

int n,h;

int main()
{
    ios::sync_with_stdio(false);
    cin >> n >> h;
    for(int i = ;i < n;i++)
    {
        cout << fixed << setprecision() << h*sqrt(1.0*i/n) << " ";
    }
    cout << endl;
    return ;
}

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C.首先可以得知，选取s1中最小的(n+1)/2个，s2中最大的n/2个，整个过程有以下两步骤。

1.当s1中最小的比s2中最小的要小是，显然，s1每次选最小的放在串首，s2每次选最小的放在串首。

2.否则，s1每次选最大的放在串尾，s2每次选最大的放在串尾。

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

string s1,s2;

int main()
{
    ios::sync_with_stdio(false);
    cin >> s1;
    sort(s1.begin(),s1.end());
    cin >> s2;
    sort(s2.begin(),s2.end());
    reverse(s2.begin(),s2.end());
    int n = (s1.length()+)/,m = s1.length()/;
    int now1 = ,now2 = ;
    char ans[] = {};
    int num = ,l = ,r = s1.length()-;
    while(num < s1.length() && s1[now1] < s2[now2])
    {
        if(num% == )  ans[l++] = s1[now1++];
        else    ans[l++] = s2[now2++];
        num++;
    }
    now1 = n-,now2 = m-;
    while(num < s1.length())
    {
        if(num% == )  ans[r--] = s1[now1--];
        else    ans[r--] = s2[now2--];
        num++;
    }
    cout << ans << endl;
    return ;
}

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D.用hash缩点，之后的图每个点最多只能有两条边，dfs。

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

int n,m;
int hashh[],pre[],ans[] = {},vis[] = {},visc[] = {};
vector <int> v[];

void dfs(int now)
{
    vis[now] = true;
    if(ans[now] == )   return;
    for(int i = ;i < v[now].size();i++)
    {
        int t = v[now][i];
        if(hashh[t] == hashh[now])  ans[t] = ans[now];
    }
    for(int i = ;i < v[now].size();i++)
    {
        int t = v[now][i];
        if(vis[t])    continue;
        if(!ans[t])
        {
            if(!visc[ans[now]-])
            {
                visc[ans[now]-] = ;
                ans[t] = ans[now]-;
            }
            else if(!visc[ans[now]+])
            {
                visc[ans[now]+] = ;
                ans[t] = ans[now]+;
            }
        }
        dfs(t);
    }
}

int main()
{
    ios::sync_with_stdio();
    cin >> n >> m;
    pre[] = ;
    for(int i = ;i <= n;i++)   pre[i] = pre[i-]*+i;
    for(int i = ;i <= n;i++)   hashh[i] = pre[i];
    for(int i = ;i <= m;i++)
    {
        int x,y;
        cin >> x >> y;
        v[x].push_back(y);
        v[y].push_back(x);
        hashh[x] += pre[y];
        hashh[y] += pre[x];
    }
    ans[] = ;
    visc[] = ;
    dfs();
    for(int i=;i<=n;i++)
    {
        if(!ans[i])
        {
            cout << "NO" << endl;
            return ;
        }
    }
    cout << "YES" << endl;
    for(int i = ;i <= n;i++)   cout << ans[i] << " ";
    cout << endl;
    return ;
}

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