Codeforces_842

A.枚举一个区间，判断是否有数符合。

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

long long l,r,x,y,k;

int main()
{
    ios::sync_with_stdio();
    cin >> l >> r >> x >> y >> k;
    for(int i = x;i <= y;i++)
    {
        if(l <= i*k && i*k <= r)
        {
            cout << "YES" << endl;
            return ;
        }
    }
    cout << "NO" << endl;
    return ;
}

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B.判断两个边界。

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

int r,d,n;

int main()
{
    ios::sync_with_stdio();
    cin >> r >> d >> n;
    int ans = ;
    for(int i = ;i <= n;i++)
    {
        double x,y,rr;
        cin >> x >> y >> rr;
        double dis = sqrt(x*x+y*y);
        if(dis-rr > r-d-1e- && dis+rr < r+1e-)    ans++;
    }
    cout << ans << endl;
    return ;
}

---

C.暴力判断每种gcd，set去重后实际的数量很少。

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

int n,a[],ans[] = {};
vector<int> v[];
set<int> s[];

void dfs(int now,int pre)
{
    for(int t : s[pre]) s[now].insert(__gcd(t,a[now]));
    a[now] = __gcd(a[pre],a[now]);
    s[now].insert(a[pre]);
    for(int t : s[now]) ans[now] = max(ans[now],t);
    for(int t : v[now])
    {
        if(t == pre)    continue;
        dfs(t,now);
    }
}

int main()
{
    ios::sync_with_stdio();
    cin >> n;
    for(int i = ;i <= n;i++)   cin >> a[i];
    for(int i = ;i < n;i++)
    {
        int x,y;
        cin >> x >> y;
        v[x].push_back(y);
        v[y].push_back(x);
    }
    a[] = ;
    s[].insert();
    dfs(,);
    for(int i = ;i <= n;i++)   cout << ans[i] << " ";
    cout << endl;
    return ;
}

---

D.前缀和，二分判断每一位，还要记录位置改变状态。

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

int n,m,cnt[] = {};

int main()
{
    ios::sync_with_stdio();
    cin >> n >> m;
    for(int i = ;i <= n;i++)
    {
        int x;
        cin >> x;
        cnt[x] = ;
    }
    for(int i = ;i <= ;i++)  cnt[i] += cnt[i-];
    int t = ;
    while(m--)
    {
        int z;
        cin >> z;
        t ^= z;
        int l = ,r = (<<)-;
        for(int i = ;i >= ;i--)
        {
            int tt = t&(<<i),mid = (l+r)/;
            if(tt)
            {
                if(cnt[r]-cnt[mid] < r-mid) l = mid+;
                else    r = mid;
            }
            else
            {
                if(cnt[mid]-(l?cnt[l-]:) < mid-l+)  r = mid;
                else    l = mid+;
            }
        }
        cout << (l^t) << endl;
    }
    return ;
}

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