Codeforces Round #629 (Div. 3) Mar/26/2020 22:35UTC+8

比赛链接 https://codeforces.com/contest/1328
比赛记录 https://blog.csdn.net/cheng__yu_/article/details/105395197

A. Divisibility Problem（水）

思路：取个余数

#include <bits/stdc++.h>
#define ll long long 
using namespace std;
int t;
ll a,b;

int main()
{
	cin>>t;
	while(t--)
	{
		cin>>a>>b;
		ll ans=b-(a%b);
		if(a%b==0)
			ans=0;
		
		cout<<ans<<endl;
	}
	return 0;
}

B. K-th Beautiful String

题意：输出字典序第 k 小的排列
思路：找规律、二分

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=2e5+10,maxm=1e5+5;
const int inf=0x7f7f7f7f,mod=998244353;
int t;
ll n,k,a[maxn];

int main()
{
    for(ll i=1;i<=maxn;++i)
        a[i]=(1+i)*i/2;
    cin>>t;
    while(t--)
    {
        cin>>n>>k;
        int p1=lower_bound(a+1,a+maxn,k)-a-1;
        int p2=k-a[p1];
        p1=n-p1-1;
        p2=n-p2+1;
        //cout<
        for(int i=1;i<=n;++i)
            if(i==p1||i==p2)
                printf("b");
            else
                printf("a");
        puts("");
    }
    return 0;
}

C. Ternary XOR

题意：把一个 3 进制的数拆分成两个数，要求最大值最小
思路：遇到 2 就平均分，遇到第一个 1 的时候分给其中一个，后面的数都分给第二个

实现：就两种状态，遇到 1 之后，改变状态，只运行一种方法就行了

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=2e5+10,maxm=1e5+5;
const int inf=0x7f7f7f7f,mod=998244353;
int t,n;
string s;

int main()
{
    cin>>t;
    while(t--)
    {
        cin>>n>>s;
        s="0"+s;
        string ans1,ans2;
        bool f=0;
        for(int i=1;i<=n;++i)
        {
            if(f)
            {
                ans2+=s[i];
                ans1+="0";
                continue;
            }
            if(s[i]=='2')
                ans1+="1",ans2+="1";
            else if(s[i]=='1')
                ans1+="1",ans2+="0",f=1;
            else
                ans1+="0",ans2+="0";
        }
        cout<<ans1<<"\n"<<ans2<<"\n";
    }
    return 0;
}

D. Carousel

题意：给 n 个动物围成一圈染色，要求相邻的不同动物不能染相同的颜色。相同动物可以随便染色

#include <bits/stdc++.h>

using namespace std;

int solve() {
	int n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	
	if (count(a.begin(), a.end(), a[0]) == n) {
		cout << 1 << endl;
		for (int i = 0; i < n; ++i) {
			cout << 1 << " ";
		}
		cout << endl;
		return 0;
	}
	
	if (n % 2 == 0) {
		cout << 2 << endl;
		for (int i = 0; i < n; ++i) {
			cout << i % 2 + 1 << " ";
		}
		cout << endl;
		return 0;
	}
	
	for (int i = 0; i < n; ++i) {
		if (a[i] == a[(i + 1) % n]) {
			vector<int> ans(n);
			for (int j = 0, pos = i + 1; pos < n; ++pos, j ^= 1) {
				ans[pos] = j + 1;
			}
			for (int j = 0, pos = i; pos >= 0; --pos, j ^= 1) {
				ans[pos] = j + 1;
			}
			cout << 2 << endl;
			for (int pos = 0; pos < n; ++pos) {
				cout << ans[pos] << " ";
			}
			cout << endl;
			return 0;
		}
	}
	
	cout << 3 << endl;
	for (int i = 0; i < n - 1; ++i) {
		cout << i % 2 + 1 << " ";
	}
	cout << 3 << endl;
    return 0;    
}

int main() {
	int q;
	cin >> q;
	for (int qq = 0; qq < q; qq++) {
	    solve();
	}
	return 0;
}

E. Tree Queries（LCA）

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=2e5+10,inf=1e9;

int head[maxn],cnt;
struct Edge
{
    int nxt,to;
}edges[maxn<<1];

void add(int u,int v)
{
    edges[++cnt].to=v;
    edges[cnt].nxt=head[u];
    head[u]=cnt;
}

int fa[maxn][25],depth[maxn];

void dfs(int u,int f)
{
    fa[u][0]=f;
    depth[u]=depth[fa[u][0]]+1;
    for(int i=1;i<=20;++i)
        fa[u][i]=fa[fa[u][i-1]][i-1];
    for(int i=head[u];i!=-1;i=edges[i].nxt)
    {
        int v=edges[i].to;
        if(v!=f) dfs(v,u);
    }
}

int lca(int u,int v)
{
    if(depth[u]>depth[v]) swap(u,v);
    int dis=depth[v]-depth[u];
    for(int j=0;dis;j++)
    {
        if(dis&1) v=fa[v][j];
        dis>>=1;
    }
    if(u==v) return u;
    for(int i=20;i>=0;--i)
    {
        if(fa[u][i]!=fa[v][i])
            u=fa[u][i],v=fa[v][i];
    }
    return fa[u][0];
}

bool cmp(int a,int b)
{
    return depth[a]<depth[b];
}

int n,m,k,v[maxn];

int main()
{
    cin>>n>>m;
    memset(head,-1,sizeof(head));
    cnt=-1;
    for(int i=1;i<=n-1;++i)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        add(u,v),add(v,u);
    }
    dfs(1,0);
    while(m--)
    {
        cin>>k;
        for(int i=1;i<=k;++i) cin>>v[i];
        sort(v+1,v+1+k,cmp);
        int mx=v[k];
        bool ok=1;
        for(int i=1;i<k;++i)
        {
            if(lca(v[i],mx)==v[i]||lca(v[i],mx)==fa[v[i]][0]) continue;
            ok=false;
            break;
        }
        puts(ok?"YES":"NO");
    }
    return 0;
}

F. Make k Equal

题意：每次只能让最大的数减 1 ，或者让最小的数加 +1，问至少需要操作几次才能够让数组有 k 个数字相等
思路：排完序后，枚举每一个位置，前后统一一下答案
$$ans=\{\begin{array}{l}a[i]\times i-{\sum }_{j=1}^{i}a[i]-(i-k)(i\ge k)\\ {\sum }_{j=i+1}^{n}a[i]-a[i]\times (n-i)-(n-i+1-k)(i+k-1\le n)\\ a[i]\times i-{\sum }_{j=1}^{i}a[i]+{\sum }_{j=i+1}^{n}a[i]-a[i]\times (n-i)-(n-k)\end{array}$$

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=2e5+10,maxm=1e5+5;
const int inf=0x7f7f7f7f,mod=998244353;
int n,k;
ll a[maxn],pref[maxn];
map<int,int> m;

int main()
{
    cin>>n>>k;
    bool ok=0;
    for(int i=1;i<=n;++i)
    {
        cin>>a[i];
        m[a[i]]++;
        if(m[a[i]]>=k)
            ok=1;
    }
    if(ok)
    {
        puts("0");
        return 0;
    }
    sort(a+1,a+1+n);
    for(int i=1;i<=n;++i)
        pref[i]=pref[i-1]+a[i];
    ll ans=9e18;
    for(int i=1;i<=n;++i)
    {
        if(i>=k)
            ans=min(ans,a[i]*i-pref[i]-(i-k));
        if(i+k-1<=n)
            ans=min(ans,pref[n]-pref[i]-a[i]*(n-i)-(n-i+1-k));
        ans=min(ans,a[i]*i-pref[i]+pref[n]-pref[i]-a[i]*(n-i)-(n-k));
    }
    cout<<ans<<"\n";
    return 0;
}