CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes)_化简|ai aj|+|aj ak|+|ak al|+|al ai|-优快云博客

本文链接：https://blog.youkuaiyun.com/nike0good/article/details/123749902

本文解析了五道算法竞赛题目，包括寻找符合条件的好对、判断序列操作的可能性、使数组元素相等的操作、寻找特定形式的表达式以及构造树的权重等。通过对每道题目的深入分析，提供了详细的解题思路和实现代码。

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A Good Pairs

You are given an array $a_1,a_2,…,a_n$ of positive integers. A good pair is a pair of indices $(i, j)$ with $1 \leq i, j \leq n$ such that, for all $1 \leq k \leq n$ , the following equality holds:
$∣ a i - a k ∣ + ∣ a k - a j ∣ = ∣ a i - a j ∣$
where $∣ x ∣$ denotes the absolute value of $x$ .

Find a good pair. Note that $i$ can be equal to $j$ .

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
pair<int,int> a[112345];
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read();
	while(T--) {
		int n=read();
		For(i,n) a[i]=mp(read(),i);
		sort(a+1,a+1+n);
		cout<<a[1].se<<' '<<a[n].se<<endl;
	}
	
	
	return 0;
}

B Subtract Operation

You are given a list of n integers. You can perform the following operation: you choose an element x from the list, erase x from the list, and subtract the value of x from all the remaining elements. Thus, in one operation, the length of the list is decreased by exactly 1.

Given an integer k (k>0), find if there is some sequence of n−1 operations such that, after applying the operations, the only remaining element of the list is equal to k.

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
set<ll> h;
int work() {
	ll n=read(),k=read();
	h.clear();
	bool b=0;
	while(n--) {
		ll a=read();
        if ( h.find(a-k)!=h.end() || h.find(a+k)!=h.end() ) {
        	b=1;
        };
        h.insert(a);
    }
    return b;
}
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read();
	while(T--) {
        puts(work()?"YES":"NO");
	}
	return 0;
}

C Make Equal With Mod

You are given an array of n non-negative integers a1,a2,…,an. You can make the following operation: choose an integer x≥2 and replace each number of the array by the remainder when dividing that number by x, that is, for all 1≤i≤n set ai to aimodx.

Determine if it is possible to make all the elements of the array equal by applying the operation zero or more times.

考虑 $ai=0a_i \bmod a_i =0$ ，所以想办法把所有数变成0，发现有1的情况不行，1不能变成其它数

分类讨论：

没有1，全变成0
有1，全变成1
情况2需要 $(ai−1)=1a_i \bmod (a_i-1) =1$ ，因此不能出现相邻的数，不能出现 $0, 2$ 。

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
int a[112345];
void work() {
		int n=read();
	For(i,n) a[i]=read();
	sort(a+1,a+1+n);
	int fl1=0;
	For(i,n) if(a[i]==1) fl1++;
	if(!fl1) puts("YES");
	else {
		For(i,n)
		if(a[i]==0||a[i]==2||(a[i]>=3&&i<n && a[i+1]==a[i]+1)) {
			puts("NO");
			return;
		}
		puts("YES");
	}
}
int main()
{
//	freopen("A.in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read();
	while(T--) {
		work();
	}
	
	
	return 0;
}

D K-good

We say that a positive integer n is k-good for some positive integer k if n can be expressed as a sum of k positive integers which give k distinct remainders when divided by k.

Given a positive integer n, find some k≥2 so that n is k-good or tell that such a k does not exist.

$k+(k+1)+⋯+(k+k−1)+tk=k(k−1)/2+tk(t≥0,t∈Z)k+(k+1)+\cdots + (k+k-1)+tk=k(k-1)/2+tk(t\ge 0,t\in Z)$
因此当k是奇数时，k是n的因子
k是偶数的情况可以覆盖所有>=3的奇数的情况

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef long long ll;
const int maxp = 1e6 + 1, maxv = 25, maxc = (int)1e4 + 1;
int ptot, pr[maxp], d[maxp], cnt;
LL n, p[maxc];
LL mod_add(LL x, LL y, LL p) {
	return (x += y) < p ? x : x - p;
}
LL mod_mul(LL x, LL y, LL p) {
	LL ret = x * y - (LL)((long double)x * y / p + 0.5) * p;
	return ret < 0 ? ret + p : ret;
}
LL mod_pow(LL x, LL k, LL p) {
	LL ret = 1 % p;
	for( ; k > 0; k >>= 1, x = mod_mul(x, x, p))
		(k & 1) && (ret = mod_mul(ret, x, p));
	return ret;
}
bool miller_rabin(LL n) {
	if(n == 2) return 1;
	if(n < 2 || !(n & 1))
		return 0;
	LL s = 0, r = n - 1;
	for( ; !(r & 1); r >>= 1, ++s);
	for(int i = 0; pr[i] < n && pr[i] < maxv; ++i) {
		LL cur = mod_pow(pr[i], r, n), nxt;
		for(int j = 0; j < s; ++j) {
			nxt = mod_mul(cur, cur, n);
			if(nxt == 1 && cur != 1 && cur != n - 1) return 0;
			cur = nxt;
		}
		if(cur != 1) return 0;
	}
	return 1;
}
LL gcd(LL a, LL b) {
	int ret = 0;
	while(a) {
		for( ; !(a & 1) && !(b & 1); ++ret, a >>= 1, b >>= 1);
		for( ; !(a & 1); a >>= 1);
		for( ; !(b & 1); b >>= 1);
		if(a < b)
			swap(a, b);
		a -= b;
	}
	return b << ret;
}
LL pollard_rho(LL n) {
	static LL seq[maxp];
	while(1) {
		LL x = rand() % n, y = x, c = rand() % n;
		LL *px = seq, *py = seq, tim = 0, prd = 1;
		while(1) {
			*py++ = y = mod_add(mod_mul(y, y, n), c, n);
			*py++ = y = mod_add(mod_mul(y, y, n), c, n);
			if((x = *px++) == y) break;
			LL tmp = prd;
			prd = mod_mul(prd, abs(y - x), n);
			if(!prd) return gcd(tmp, n);
			if((++tim) == maxv) {
				if((prd = gcd(prd, n)) > 1 && prd < n) return prd;
				tim = 0;
			}
		}
		if(tim && (prd = gcd(prd, n)) > 1 && prd < n) return prd;
	}
}
void decompose(LL n) {
	for(int i = 0; i < cnt; ++i)
		if(n % p[i] == 0) {
			p[cnt++] = p[i];
			n /= p[i];
		}
	if(n < maxp) {
		for( ; n > 1; p[cnt++] = d[n], n /= d[n]);
	} else if(miller_rabin(n)) {
		p[cnt++] = n;
	} else {
		LL fact = pollard_rho(n);
		decompose(fact), decompose(n / fact);
	}
} // prepare pr(prime) and d(minimal factor)
map<LL,int> h;
ll xs[3000];
ll xs2[3000];
vector<vector<LL> > table;
int main() {
	for(int i = 2; i < maxp; ++i) {
		if(!d[i])
			pr[ptot++] = d[i] = i;
		for(int j = 0, k; (k = i * pr[j]) < maxp; ++j) {
			d[k] = pr[j];
			if(d[i] == pr[j])
				break;
		}
	}
	int m, mod = 1000000007;
	int T;cin>>T;
	while(T--) {
		cnt=0;
		LL n;
		cin>>n;
		LL n2=n;
		LL b = n;
		int a = 0;
		LL pa = 1;
	    while ((b & 1) == 0) {
	        b /= 2;
			pa *= 2;
	    }
	    pa *= 2;
	    if (pa + 1 <= b) {
			cout<<pa<<endl; continue;
		}
		decompose(n);
		sort(p, p + cnt);
		ll ans=-1;
		for(int i=0;i<cnt;i++) if(p[i]!=2){
			bool fl=2.0*n2/p[i]>=(p[i]-1);
			if (fl) {
				ans=p[i];break;
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}

E Equal Tree Sums

You are given an undirected unrooted tree, i.e. a connected undirected graph without cycles.

You must assign a nonzero integer weight to each vertex so that the following is satisfied: if any vertex of the tree is removed, then each of the remaining connected components has the same sum of weights in its vertices.

 #include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i>0;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,cal) printf("Case #%d: %lld\n",kcase,cal);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define add(a,b) ((a+b)%F)
ll mul(ll a,ll b){return (a*b)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 

#define MAXN (212345)
ll n;
vi v[MAXN];
int ans[MAXN];
void dfs(int x,int fa,int fl) {
	ans[x]=v[x].size();
	if(fl)ans[x]*=-1;
	for (auto u:v[x])
		if (u!=fa) {
			dfs(u,x,fl^1);
	}
}
int main()
{
//	freopen("c.in","r",stdin);
//	freopen(".out",w",stdout);
	int T=read();
	while(T--) {
		n=read();
		For(i,n-1) {
			int x=read(),y=read();
			v[x].pb(y); v[y].pb(x);
		}
		dfs(1,-1,1);
		For(i,n) v[i].resize(0);
		PRi(ans,n)
	}
	return 0;
}

F Parametric MST

You are given $n$ integers $a 1, a 2, \dots, a n$ . For any real number t, consider the complete weighted graph on $n$ vertices $K n (t)$ with weight of the edge between vertices $i$ and $j$ equal to $w_{ij}(t)=a_i⋅a_j+t⋅(a_i+a_j)$ .

Let $f (t)$ be the cost of the minimum spanning tree of $K n (t)$ . Determine whether $f (t)$ is bounded above and, if so, output the maximum value it attains.

Input
The input consists of multiple test cases. The first line contains a single integer $T (1≤T≤10^4)$ — the number of test cases. Description of the test cases follows.

The first line of each test case contains an integer $n (2≤n≤2⋅10^5)$ — the number of vertices of the graph.

The second line of each test case contains $n$ integers $a_1,a_2,…,a_n (−10^6 ≤ai≤ 10^6)$ .

The sum of n for all test cases is at most $2⋅10^5$ .

Output
For each test case, print a single line with the maximum value of $f (t)$ (it can be shown that it is an integer), or INF if $f (t)$ is not bounded above.

Example
inputCopy
5
2
1 0
2
-1 1
3
1 -1 -2
3
3 -1 -2
4
1 2 3 -4
outputCopy
INF
-1
INF
-6
-18

$w(i,j)=w_{ij}(t)=a_i⋅a_j+t⋅(a_i+a_j) = (a_i+t)(a_j+t)-t^2$
考虑最小生成树，对于点 $i$ 而言，当 $a_i+t<0$ 时连 $a_j+t$ 最大的，当 $a_i+t>0$ 时连 $a_j+t$ 最小的，当 $a_i+t=0$ 时连任意一个。

最后考察当 $t$ 为正、负无穷时，最小生成树边权和是否为正无穷即可。

 #include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i>0;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,cal) printf("Case #%d: %lld\n",kcase,cal);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define add(a,b) ((a+b)%F)
ll mul(ll a,ll b){return (a*b)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 

#define MAXN (212345)
ll n;
ll a[MAXN];
bool fl=0;
ll work() {
	ll ans=-1e18;
	sort(a+1,a+1+n);
	ll B=0,K=0;
	Fork(i,2,n) K+=a[1]+a[i];
	if(K>0){ // means no upbound when t=+oo
		fl=1;return 0;
	} 
	K=0;
	For(i,n-1) K+=a[i]+a[n];
	if(K<0) {// means no upbound when t=-oo
		fl=1;return 0;
	}
	
	K=0;	
	Fork(i,2,n) K+=a[1]+a[i],B+=a[1]*a[i];
	// 1-n 2-n k-n 1-(k+1) 1-n-1
		
	gmax(ans,K*max(-a[1],-a[n])+B);
	
	Fork(i,2,n-1) {
		K+=-a[1]+a[n];B+=-a[1]*a[i]+a[i]*a[n];
		gmax(ans,K*(-a[i+1])+B);
		gmax(ans,K*(-a[i])+B);
	}
	return ans;	
}
int main()
{
//	freopen("c.in","r",stdin);
//	freopen(".out",w",stdout);
	int T=read();
	while(T--) {
		n=read();
		For(i,n) a[i]=read();
		fl=0;
		ll t=work();
		if(fl) puts("INF");else cout<<t<<endl;
	}
	return 0;
}