# Codeforces Round #690 (Div. 3) 题解（A-E2)

2020-12-17 23:05:48  阅读：102  来源： 互联网

# A. Favorite Sequence

## 题目

Polycarp has a favorite sequence a[1…n] consisting of n integers. He wrote it out on the whiteboard as follows:

he wrote the number a1 to the left side (at the beginning of the whiteboard);
he wrote the number a2 to the right side (at the end of the whiteboard);
then as far to the left as possible (but to the right from a1), he wrote the number a3;
then as far to the right as possible (but to the left from a2), he wrote the number a4;
Polycarp continued to act as well, until he wrote out the entire sequence on the whiteboard.

For example, if n=7 and a=[3,1,4,1,5,9,2], then Polycarp will write a sequence on the whiteboard [3,4,5,2,9,1,1].

You saw the sequence written on the whiteboard and now you want to restore Polycarp's favorite sequence.

## input & output

Input
The first line contains a single positive integer t (1≤t≤300) — the number of test cases in the test. Then t test cases follow.

The first line of each test case contains an integer n (1≤n≤300) — the length of the sequence written on the whiteboard.

The next line contains n integers b1,b2,…,bn (1≤bi≤109) — the sequence written on the whiteboard.

Output
Output t answers to the test cases. Each answer — is a sequence a that Polycarp wrote out on the whiteboard.

Example
inputCopy
6
7
3 4 5 2 9 1 1
4
9 2 7 1
11
8 4 3 1 2 7 8 7 9 4 2
1
42
2
11 7
8
1 1 1 1 1 1 1 1
outputCopy
3 1 4 1 5 9 2
9 1 2 7
8 2 4 4 3 9 1 7 2 8 7
42
11 7
1 1 1 1 1 1 1 1
Note
In the first test case, the sequence a matches the sequence from the statement. The whiteboard states after each step look like this:

[3]⇒[3,1]⇒[3,4,1]⇒[3,4,1,1]⇒[3,4,5,1,1]⇒[3,4,5,9,1,1]⇒[3,4,5,2,9,1,1].

## 代码

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

const int maxn = 1000+10;

int a[maxn] , b[maxn];
int main()
{
int t;
cin>>t;
while(t--)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
int n;
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i];
int l = 1 , r = n , sum = n , now = 1;
int f = 1;
while(sum--)
{
if(f)
b[now++] = a[l] , l++ , f = 0;
else
b[now++] = a[r] , r-- , f = 1;
}
for(int i=1;i<=n;i++) cout<<b[i]<<" ";
cout<<"\n";
}
}
``````

# B. Last Year's Substring

## 题目

Polycarp has a string s[1…n] of length n consisting of decimal digits. Polycarp performs the following operation with the string s no more than once (i.e. he can perform operation 0 or 1 time):

Polycarp selects two numbers i and j (1≤i≤j≤n) and removes characters from the s string at the positions i,i+1,i+2,…,j (i.e. removes substring s[i…j]). More formally, Polycarp turns the string s into the string s1s2…si−1sj+1sj+2…sn.
For example, the string s="20192020" Polycarp can turn into strings:

"2020" (in this case (i,j)=(3,6) or (i,j)=(1,4));
"2019220" (in this case (i,j)=(6,6));
"020" (in this case (i,j)=(1,5));
other operations are also possible, only a few of them are listed above.
Polycarp likes the string "2020" very much, so he is wondering if it is possible to turn the string s into a string "2020" in no more than one operation? Note that you can perform zero operations.

## input & output

Input
The first line contains a positive integer t (1≤t≤1000) — number of test cases in the test. Then t test cases follow.

The first line of each test case contains an integer n (4≤n≤200) — length of the string s. The next line contains a string s of length n consisting of decimal digits. It is allowed that the string s starts with digit 0.

Output
For each test case, output on a separate line:

"YES" if Polycarp can turn the string s into a string "2020" in no more than one operation (i.e. he can perform 0 or 1 operation);
"NO" otherwise.
You may print every letter of "YES" and "NO" in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Example
inputCopy
6
8
20192020
8
22019020
4
2020
5
20002
6
729040
6
200200
outputCopy
YES
YES
YES
NO
NO
NO
Note
In the first test case, Polycarp could choose i=3 and j=6.

In the second test case, Polycarp could choose i=2 and j=5.

In the third test case, Polycarp did not perform any operations with the string.

2020在开头
2020在结尾
202XXX0
20XXX20
2XXX020

## 代码

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

int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
string s;
cin>>s;
int len = s.size();
if(len<4)
{
cout<<"NO\n";
continue;
}
else
{
if(s[0]=='2' && s[1]=='0' && s[2]=='2' && s[3] =='0')
{
cout<<"YES\n";
continue;
}
else if(s[len-4]=='2' && s[len-3]=='0' && s[len-2]=='2' && s[len-1] == '0')
{
cout<<"YES\n";
continue;
}
else if(s[0]=='2' && s[len-3]=='0' && s[len-2]=='2' && s[len-1] == '0')
{
cout<<"YES\n";
continue;
}
else if(s[0]=='2' && s[1]=='0' && s[len-2]=='2' && s[len-1] == '0')
{
cout<<"YES\n";
continue;
}
else if(s[0]=='2' && s[1]=='0' && s[2]=='2' && s[len-1] == '0')
{
cout<<"YES\n";
continue;
}
else
cout<<"NO\n";
}
}
}
``````

# C. Unique Number

## 题目

You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).

## input & output

Input
The first line contains a single positive integer t (1≤t≤50) — the number of test cases in the test. Then t test cases follow.

Each test case consists of a single integer number x (1≤x≤50).

Output
Output t answers to the test cases:

if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number;
otherwise print -1.
Example
inputCopy
4
1
5
15
50
outputCopy
1
5
69
-1

## 代码

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

int main()
{
int t;
cin>>t;
while(t--)
{
int x;
cin>>x;
if(x<=9)
cout<<x<<"\n";
else
{
if(x<=17)
cout<<x-9<<""<<9<<"\n";
else if(x<=24)
cout<<x-17<<""<<8<<""<<9<<"\n";
else if(x<=30)
cout<<x-24<<""<<7<<""<<8<<""<<9<<"\n";
else if(x<=35)
cout<<x-30<<""<<6<<""<<7<<""<<8<<""<<9<<"\n";
else if(x<=39)
cout<<x-35<<""<<5<<""<<6<<""<<7<<""<<8<<""<<9<<"\n";
else if(x<=42)
cout<<x-39<<""<<4<<""<<5<<""<<6<<""<<7<<""<<8<<""<<9<<"\n";
else if(x<=44)
cout<<x-42<<""<<3<<""<<4<<""<<5<<""<<6<<""<<7<<""<<8<<""<<9<<"\n";
else if(x<=45)
cout<<x-44<<""<<2<<""<<3<<""<<4<<""<<5<<""<<6<<""<<7<<""<<8<<""<<9<<"\n";
else
cout<<-1<<"\n";
}
}
}
``````

# D. Add to Neighbour and Remove

## 题目

Polycarp was given an array of a[1…n] of n integers. He can perform the following operation with the array a no more than n times:

Polycarp selects the index i and adds the value ai to one of his choice of its neighbors. More formally, Polycarp adds the value of ai to ai−1 or to ai+1 (if such a neighbor does not exist, then it is impossible to add to it).
After adding it, Polycarp removes the i-th element from the a array. During this step the length of a is decreased by 1.
The two items above together denote one single operation.

For example, if Polycarp has an array a=[3,1,6,6,2], then it can perform the following sequence of operations with it:

Polycarp selects i=2 and adds the value ai to (i−1)-th element: a=[4,6,6,2].
Polycarp selects i=1 and adds the value ai to (i+1)-th element: a=[10,6,2].
Polycarp selects i=3 and adds the value ai to (i−1)-th element: a=[10,8].
Polycarp selects i=2 and adds the value ai to (i−1)-th element: a=[18].
Note that Polycarp could stop performing operations at any time.

Polycarp wondered how many minimum operations he would need to perform to make all the elements of a equal (i.e., he wants all ai are equal to each other).

## input & output

Input
The first line contains a single integer t (1≤t≤3000) — the number of test cases in the test. Then t test cases follow.

The first line of each test case contains a single integer n (1≤n≤3000) — the length of the array. The next line contains n integers a1,a2,…,an (1≤ai≤105) — array a.

It is guaranteed that the sum of n over all test cases does not exceed 3000.

Output
For each test case, output a single number — the minimum number of operations that Polycarp needs to perform so that all elements of the a array are the same (equal).

Example
inputCopy
4
5
3 1 6 6 2
4
1 2 2 1
3
2 2 2
4
6 3 2 1
outputCopy
4
2
0
2
Note
In the first test case of the example, the answer can be constructed like this (just one way among many other ways):

[3,1,6,6,2] −→−−−−−−−i=4, add to left [3,1,12,2] −→−−−−−−−−i=2, add to right [3,13,2] −→−−−−−−−−i=1, add to right [16,2] −→−−−−−−−i=2, add to left [18]. All elements of the array [18] are the same.

In the second test case of the example, the answer can be constructed like this (just one way among other ways):

[1,2,2,1] −→−−−−−−−−i=1, add to right [3,2,1] −→−−−−−−−i=3, add to left [3,3]. All elements of the array [3,3] are the same.

In the third test case of the example, Polycarp doesn't need to perform any operations since [2,2,2] contains equal (same) elements only.

In the fourth test case of the example, the answer can be constructed like this (just one way among other ways):

[6,3,2,1] −→−−−−−−−−i=3, add to right [6,3,3] −→−−−−−−−i=3, add to left [6,6]. All elements of the array [6,6] are the same.

## 代码

``````#include <bits/stdc++.h>
using namespace std;
#define xiaobai false
#define plog if(xiaobai) clog

const int maxn = 3000+100;
int a[maxn];
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
long long sum = 0;
cin>>n;
for(int i=1;i<=n;i++) cin>>a[i] , sum += a[i];
int ans = 0 , flag = 0;
for(int i=n;i>=1;i--)
{
if(sum % i == 0)
{
//				cout<<sum/i<<endl;
long long cnt = 0;
for(int j=1;j<=n;j++)
{
cnt += a[j];
plog<<"before: "<<cnt<<endl;
if(cnt==sum/i)
cnt=0;
plog<<"after: "<<cnt<<endl;
}
if(cnt == 0)
{
ans = n - i;
flag = 1;
}
}
if(flag)
break;
}
cout<<ans<<"\n";
}
}
``````

# E2. Close Tuples (hard version)

## 题目

This is the hard version of this problem. The only difference between the easy and hard versions is the constraints on k and m. In this version of the problem, you need to output the answer by modulo 109+7.

You are given a sequence a of length n consisting of integers from 1 to n. The sequence may contain duplicates (i.e. some elements can be equal).

Find the number of tuples of m elements such that the maximum number in the tuple differs from the minimum by no more than k. Formally, you need to find the number of tuples of m indices i1<i2<…<im, such that

max(ai1,ai2,…,aim)−min(ai1,ai2,…,aim)≤k.
For example, if n=4, m=3, k=2, a=[1,2,4,3], then there are two such triples (i=1,j=2,z=4 and i=2,j=3,z=4). If n=4, m=2, k=1, a=[1,1,1,1], then all six possible pairs are suitable.

As the result can be very large, you should print the value modulo 109+7 (the remainder when divided by 109+7).

## input & output

Input
The first line contains a single integer t (1≤t≤2⋅105) — the number of test cases. Then t test cases follow.

The first line of each test case contains three integers n, m, k (1≤n≤2⋅105, 1≤m≤100, 1≤k≤n) — the length of the sequence a, number of elements in the tuples and the maximum difference of elements in the tuple.

The next line contains n integers a1,a2,…,an (1≤ai≤n) — the sequence a.

It is guaranteed that the sum of n for all test cases does not exceed 2⋅105.

Output
Output t answers to the given test cases. Each answer is the required number of tuples of m elements modulo 109+7, such that the maximum value in the tuple differs from the minimum by no more than k.

Example
inputCopy
4
4 3 2
1 2 4 3
4 2 1
1 1 1 1
1 1 1
1
10 4 3
5 6 1 3 2 9 8 1 2 4
outputCopy
2
6
1
20

## 代码

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

typedef long long ll;
#define xiaobai false
#define plog if(xiaobai) clog

const ll MOD = 1e9+7;
const int MAXN= 2e5+5;

const int maxn=2e5+5;
const ll mod=1e9+7;
ll fac[maxn],inv[maxn];
ll pow_mod(ll a,ll n)
{
ll ret =1;
while(n)
{
if(n&1) ret=ret*a%mod;
a=a*a%mod;
n>>=1;
}
return ret;
}
void init()
{
fac[0]=1;
for(int i=1;i<maxn;i++)
{
fac[i]=fac[i-1]*i%mod;
}
}
ll Cc(ll x, ll y)
{
return fac[x]*pow_mod(fac[y]*fac[x-y]%mod,mod-2)%mod;
}

ll a[MAXN];
int main()
{
//	deque<ll>q;
init();
int t;
cin>>t;
while(t--)
{
deque<ll>q;
int n,m,k;
cin>>n>>m>>k;
ll ans = 0;
for(int i=1;i<=n;i++) cin>>a[i];
sort(a+1,a+1+n);
for(int i=1;i<=n;i++) plog<<a[i]<<" ";
plog<<endl;
if(m==1)
{
cout<<n<<"\n";
continue;
}
for(int i=1;i<=n;i++)
{
q.push_back(a[i]);
while(!q.empty() && q.back() - q.front() > k) //如果最后面的-前面的>k 就pop前面的
q.pop_front();
if(q.size() >= m)
{
ans += Cc(q.size()-1,m-1); // 假设最后一位是结尾，固定后可以去重
ans %= MOD;
}
}
cout<<ans<<"\n";
}
}
``````

# E1. Close Tuples (easy version)

## 题目

This is the easy version of this problem. The only difference between easy and hard versions is the constraints on k and m (in this version k=2 and m=3). Also, in this version of the problem, you DON'T NEED to output the answer by modulo.

You are given a sequence a of length n consisting of integers from 1 to n. The sequence may contain duplicates (i.e. some elements can be equal).

Find the number of tuples of m=3 elements such that the maximum number in the tuple differs from the minimum by no more than k=2. Formally, you need to find the number of triples of indices i<j<z such that

max(ai,aj,az)−min(ai,aj,az)≤2.
For example, if n=4 and a=[1,2,4,3], then there are two such triples (i=1,j=2,z=4 and i=2,j=3,z=4). If n=4 and a=[1,1,1,1], then all four possible triples are suitable.

## input & output

Input
The first line contains a single integer t (1≤t≤2⋅105) — the number of test cases. Then t test cases follow.

The first line of each test case contains an integer n (1≤n≤2⋅105) — the length of the sequence a.

The next line contains n integers a1,a2,…,an (1≤ai≤n) — the sequence a.

It is guaranteed that the sum of n for all test cases does not exceed 2⋅105.

Output
Output t answers to the given test cases. Each answer is the required number of triples of elements, such that the maximum value in the triple differs from the minimum by no more than 2. Note that in difference to the hard version of the problem, you don't need to output the answer by modulo. You must output the exact value of the answer.

Example
inputCopy
4
4
1 2 4 3
4
1 1 1 1
1
1
10
5 6 1 3 2 9 8 1 2 4
outputCopy
2
4
0
15

## 代码

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

typedef long long ll;
#define xiaobai false
#define plog if(xiaobai) clog

const int MAXN= 2e5+5;

const int maxn=2e5+5;

long long cal( long long n, long long m) {
long long i, a, b, p;
if (n<m) {
i=m;
m=n;
n=i;
}
p=1;
a=n-m<m?n-m:m;
b=n-m>m?n-m:m;
for (i=1; i<=a; i++)
p+=(p*b/i);
return p;
}

ll a[MAXN];
int main()
{
//	deque<ll>q;
int t;
cin>>t;
while(t--)
{
deque<ll>q;
int n,m,k;
cin>>n;
m = 3;
k = 2;
ll ans = 0;
for(int i=1;i<=n;i++) cin>>a[i];
sort(a+1,a+1+n);
for(int i=1;i<=n;i++) plog<<a[i]<<" ";
plog<<endl;
for(int i=1;i<=n;i++)
{
q.push_back(a[i]);
while(!q.empty() && q.back() - q.front() > k) //如果最后面的-前面的>k 就pop前面的
q.pop_front();
if(q.size() >= m)
{
ans += cal(q.size()-1,m-1); // 假设最后一位是结尾，固定后可以去重
}
}
cout<<ans<<"\n";
}
}
``````