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 = 0;
    while(n--)
    {
        int x;
        cin >> x;
        if(b < x && x < c)  ans++;
    }
    cout << ans << endl;
    return 0;
}

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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 = 1;i < n;i++)
    {
        cout << fixed << setprecision(10) << h*sqrt(1.0*i/n) << " ";
    }
    cout << endl;
    return 0;
}

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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()+1)/2,m = s1.length()/2;
    int now1 = 0,now2 = 0;
    char ans[300005] = {0};
    int num = 0,l = 0,r = s1.length()-1;
    while(num < s1.length() && s1[now1] < s2[now2])
    {
        if(num%2 == 0)  ans[l++] = s1[now1++];
        else    ans[l++] = s2[now2++];
        num++;
    }
    now1 = n-1,now2 = m-1;
    while(num < s1.length())
    {
        if(num%2 == 0)  ans[r--] = s1[now1--];
        else    ans[r--] = s2[now2--];
        num++;
    }
    cout << ans << endl;
    return 0;
}

---

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[300005],pre[300005],ans[300005] = {0},vis[300005] = {0},visc[600005] = {0};
vector <int> v[300005];

void dfs(int now)
{
    vis[now] = true;
    if(ans[now] == 0)   return;
    for(int i = 0;i < v[now].size();i++)
    {
        int t = v[now][i];
        if(hashh[t] == hashh[now])  ans[t] = ans[now];
    }
    for(int i = 0;i < v[now].size();i++)
    {
        int t = v[now][i];
        if(vis[t])    continue;
        if(!ans[t])
        {
            if(!visc[ans[now]-1])
            {
                visc[ans[now]-1] = 1;
                ans[t] = ans[now]-1;
            }
            else if(!visc[ans[now]+1])
            {
                visc[ans[now]+1] = 1;
                ans[t] = ans[now]+1;
            }
        }
        dfs(t);
    }
}

int main()
{
    ios::sync_with_stdio(0);
    cin >> n >> m;
    pre[0] = 0;
    for(int i = 1;i <= n;i++)   pre[i] = pre[i-1]*31+i;
    for(int i = 1;i <= n;i++)   hashh[i] = pre[i];
    for(int i = 1;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[1] = 300000;
    visc[300000] = 1;
    dfs(1);
    for(int i=1;i<=n;i++)
    {
        if(!ans[i])
        {
            cout << "NO" << endl;
            return 0;
        }
    }
    cout << "YES" << endl;
    for(int i = 1;i <= n;i++)   cout << ans[i] << " ";
    cout << endl;
    return 0;
}

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