CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes) A-B
C之后的尚未看题解
A：
A. Good Pairs
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given an array a1,a2,…,ana1,a2,…,an of positive integers. A good pair is a pair of indices (i,j)(i,j) with 1≤i,j≤n1≤i,j≤n such that, for all 1≤k≤n1≤k≤n, the following equality holds:
|ai−ak|+|ak−aj|=|ai−aj|,|ai−ak|+|ak−aj|=|ai−aj|,
where |x||x| denotes the absolute value of xx.
Find a good pair. Note that ii can be equal to jj.
Input
The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤10001≤t≤1000) — the number of test cases. Description of the test cases follows.
The first line of each test case contains an integer nn (1≤n≤1051≤n≤105) — the length of the array.
The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) where aiai is the ii-th element of the array.
The sum of nn for all test cases is at most 2⋅1052⋅105.
Output
For each test case, print a single line with two space-separated indices ii and jj which form a good pair of the array. The case i=ji=j is allowed. It can be shown that such a pair always exists. If there are multiple good pairs, print any of them.
Example
input
Copy
3 3 5 2 7 5 1 4 2 2 3 1 2
output
Copy
2 3 1 2 1 1
Note
In the first case, for i=2i=2 and j=3j=3 the equality holds true for all kk:
- k=1k=1: |a2−a1|+|a1−a3|=|2−5|+|5−7|=5=|2−7|=|a2−a3||a2−a1|+|a1−a3|=|2−5|+|5−7|=5=|2−7|=|a2−a3|,
- k=2k=2: |a2−a2|+|a2−a3|=|2−2|+|2−7|=5=|2−7|=|a2−a3||a2−a2|+|a2−a3|=|2−2|+|2−7|=5=|2−7|=|a2−a3|,
- k=3k=3: |a2−a3|+|a3−a3|=|2−7|+|7−7|=5=|2−7|=|a2−a3||a2−a3|+|a3−a3|=|2−7|+|7−7|=5=|2−7|=|a2−a3|.
#include
using namespace std;
typedef long long ll;
int t,n,p,q;
const int maxn=1e5+5;
ll a[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin>>t;
while(t--)
{
int mmax=0;
int mmin=1e9+5;
cin>>n;
for(int i=1; i<=n; i++)
{
cin>>a[i];
if(a[i]>mmax)
{
mmax=a[i];
p=i;
}
if(a[i] B：
B. Subtract Operation
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a list of nn integers. You can perform the following operation: you choose an element xx from the list, erase xx from the list, and subtract the value of xx from all the remaining elements. Thus, in one operation, the length of the list is decreased by exactly 11.
Given an integer kk (k>0k>0), find if there is some sequence of n−1n−1 operations such that, after applying the operations, the only remaining element of the list is equal to kk.
Input
The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases. Description of the test cases follows.
The first line of each test case contains two integers nn and kk (2≤n≤2⋅1052≤n≤2⋅105, 1≤k≤1091≤k≤109), the number of integers in the list, and the target value, respectively.
The second line of each test case contains the nn integers of the list a1,a2,…,ana1,a2,…,an (−109≤ai≤109−109≤ai≤109).
It is guaranteed that the sum of nn over all test cases is not greater that 2⋅1052⋅105.
Output
For each test case, print YES if you can achieve kk with a sequence of n−1n−1 operations. Otherwise, print NO.
You may print each letter in any case (for example, "YES", "Yes", "yes", "yEs" will all be recognized as a positive answer).
Example
input
Copy
4 4 5 4 2 2 7 5 4 1 9 1 3 4 2 17 17 0 2 17 18 18
output
Copy
YES NO YES NO
Note
In the first example we have the list {4,2,2,7}{4,2,2,7}, and we have the target k=5k=5. One way to achieve it is the following: first we choose the third element, obtaining the list {2,0,5}{2,0,5}. Next we choose the first element, obtaining the list {−2,3}{−2,3}. Finally, we choose the first element, obtaining the list {5}{5}.
#include
using namespace std;
typedef long long ll;
ll t,n,k;
const int maxn=2e5+5;
ll a[maxn];
int lowbou(ll a[],ll l,ll r,ll tar)
{
while(l<=r)
{
ll mid=(l+r)/2;
if (a[mid]>t;
while(t--)
{
memset(a,0,sizeof(a));
cin>>n>>k;
bool flag=0;
for(ll i=1; i<=n; i++)
{
cin>>a[i];
}
sort(a+1,a+1+n);
for(ll i=1; i<=n; i++)
{
ll y=lowbou(a,1,n,a[i]+k);
if(y<=n)
{
if(a[y]==a[i]+k)
{
cout<<"YES"< 关于b题的两个坑点，一开始我没有初始化memset(a,0,sizeof(a)),导致指针到了n+1还有数，初始化以后，还是错，原因在于比如2 -2 -2 -1这组数据，我-2一直搜，最后加上0刚好是-2，所以指针不能大于n，也就是必须小于等于n。题解用的是set，我看有map，上完课，回来补