CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes)
Posted nike0good
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes)相关的知识，希望对你有一定的参考价值。
A Good Pairs
You are given an array
a
1
,
a
2
,
…
,
a
n
a_1,a_2,…,a_n
a1,a2,…,an of positive integers. A good pair is a pair of indices
(
i
,
j
)
(i,j)
(i,j) with
1
≤
i
,
j
≤
n
1≤i,j≤n
1≤i,j≤n such that, for all
1
≤
k
≤
n
1≤k≤n
1≤k≤n, the following equality holds:
∣
a
i
−
a
k
∣
+
∣
a
k
−
a
j
∣
=
∣
a
i
−
a
j
∣
|ai−ak|+|ak−aj|=|ai−aj|
∣ai−ak∣+∣ak−aj∣=∣ai−aj∣
where
∣
x
∣
|x|
∣x∣ denotes the absolute value of
x
x
x.
Find a good pair. Note that i i i can be equal to j j 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以上是关于CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes)的主要内容，如果未能解决你的问题，请参考以下文章
CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes)
CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes)
CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes)
CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes)（ABCDE）