Codeforces Round #440

A. Search for Pretty Integers

You are given two lists of non-zero digits.

Let’s call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?

Input
The first line contains two integers n and m (1 ≤ n, m ≤ 9) — the lengths of the first and the second lists, respectively.

The second line contains n distinct digits a1, a2, …, an (1 ≤ ai ≤ 9) — the elements of the first list.

The third line contains m distinct digits b1, b2, …, bm (1 ≤ bi ≤ 9) — the elements of the second list.

Output
Print the smallest pretty integer.

Examples
input
2 3
4 2
5 7 6
output
25
input
8 8
1 2 3 4 5 6 7 8
8 7 6 5 4 3 2 1
output
1
Note
In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don’t have digits from the second list.

In the second example all integers that have at least one digit different from 9 are pretty. It’s obvious that the smallest among them is 1, because it’s the smallest positive integer.

这道题是问你给你两个数列，让你一个数字，同时包括两个数列中的某个数，要注意的是比如1 9和2 9这种只要9就可以，不用12

#include<bits/stdc++.h>using namespace std;const int INF=0x3f3f3f3f;bool a[20],b[20];int main(){    int n,m,x;    cin>>n>>m;    memset(a,false,sizeof(a));    memset(b,false,sizeof(b));    for(int i=1;i<=n;i++) {        cin>>x;        a[x]=true;    }    for(int i=1;i<=m;i++) {        cin>>x;        b[x]=true;    }    int ans=INF;    for(int i=1;i<=9;i++) {        if(a[i]==true&&b[i]==true) ans=min(ans,i);    }    for(int i=1;i<=9;i++) {        for(int j=i+1;j<=9;j++) {            if(a[i]==true&&b[j]==true) {                ans=min(ans,i*10+j);            }        }    }    for(int i=1;i<=9;i++) {        for(int j=i+1;j<=9;j++) {            if(b[i]==true&&a[j]==true) {                ans=min(ans,i*10+j);            }        }    }    cout<<ans<<endl;    return 0;}

B. Maximum of Maximums of Minimums

You are given an array a1, a2, …, an consisting of n integers, and an integer k. You have to split the array into exactly k non-empty subsegments. You’ll then compute the minimum integer on each subsegment, and take the maximum integer over the k obtained minimums. What is the maximum possible integer you can get?

Definitions of subsegment and array splitting are given in notes.

Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤  105) — the size of the array a and the number of subsegments you have to split the array to.

The second line contains n integers a1,  a2,  …,  an ( - 109  ≤  ai ≤  109).

Output
Print single integer — the maximum possible integer you can get if you split the array into k non-empty subsegments and take maximum of minimums on the subsegments.

Examples
input
5 2
1 2 3 4 5
output
5
input
5 1
-4 -5 -3 -2 -1
output
-5
Note
A subsegment [l,  r] (l ≤ r) of array a is the sequence al,  al + 1,  …,  ar.

Splitting of array a of n elements into k subsegments [l1, r1], [l2, r2], …, [lk, rk] (l1 = 1, rk = n, li = ri - 1 + 1 for all i > 1) is k sequences (al1, …, ar1), …, (alk, …, ark).

In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are min(1, 2, 3, 4) = 1 and min(5) = 5. The resulting maximum is max(1, 5) = 5. It is obvious that you can’t reach greater result.

In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4,  - 5,  - 3,  - 2,  - 1). The only minimum is min( - 4,  - 5,  - 3,  - 2,  - 1) =  - 5. The resulting maximum is  - 5.

这道题是问你给你一列数，问你能分成k个区间的，问每个区间的最小值中的最大值，这道题要区分k来判断，如果k==1，就是求区间最小值，如果k==2，就是求第一个数和最后一个数的最大值，因为只有不管区间内是怎么变化，肯定是从第一个和第二个之间或最后一个和倒数第二个之间断开使最小值最大，k==3时就是求区间最大值

#include<bits/stdc++.h>using namespace std;const int INF=0x3f3f3f3f;const int maxn=1e5+5;int ans[maxn];int main(){    int n,k,sum;    cin>>n>>k;    for(int i=1;i<=n;i++) cin>>ans[i];    if(k==1) {        sum=ans[1];        for(int i=2;i<=n;i++) sum=min(sum,ans[i]);    }    else if(k==2) {        sum=max(ans[1],ans[n]);    }    else {        sum=ans[1];        for(int i=2;i<=n;i++) sum=max(sum,ans[i]);    }    cout<<sum<<endl;    return 0;}

C. Maximum splitting

You are given several queries. In the i-th query you are given a single positive integer ni. You are to represent ni as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.

An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.

Input
The first line contains single integer q (1 ≤ q ≤ 105) — the number of queries.

q lines follow. The (i + 1)-th line contains single integer ni (1 ≤ ni ≤ 109) — the i-th query.

Output
For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.

Examples
input
1
12
output
3
input
2
6
8
output
1
2
input
3
1
2
3
output
-1
-1
-1
Note
12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.

8 = 4 + 4, 6 can’t be split into several composite summands.

1, 2, 3 are less than any composite number, so they do not have valid splittings.

题意是给你一个数n，问你最多可以将n分解成多少个合数的和，因为分解成一个合数只能分解成4，6，9，否则就不是最大的，然后现在有四种情况：
1.n%4==0全部分解成4
2.n%4==1，可以分解成4+4+1和其他的4
3.n%4==2 可以分解成4+2和其他的4
4.n%4==3 可以分解成4+4+1和4+2和其他的4

#include<bits/stdc++.h>using namespace std;#define endl '\n'int main(){    ios::sync_with_stdio(0);    cin.tie(0);    int T;    cin>>T;    while(T--) {        int n,ans;        cin>>n;        if(n%4==0) ans=n/4;        else if(n%4==1) {            if(n/4>=2) ans=n/4-1;            else ans=-1;        }        else if(n%4==2) {            if(n/4>=1) ans=n/4;            else ans=-1;        }        else if(n%4==3) {            if(n/4>=3) ans=n/4-1;            else ans=-1;        }        cout<<ans<<endl;    }    return 0;}