Codeforces Round #486 (Div. 3)

A. Diverse Team

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are $\mathrm{n students\; in\; a\; school\; class,\; the\; rating\; of\; the ii-th\; student\; on\; Codehorses\; is ai. You\; have\; to\; form\; a\; team\; consisting\; of k students\; (1\le k\le n)\; such\; that\; the\; ratings\; of\; all\; team\; members are\; distinct.}$

If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $\mathrm{kk distinct\; numbers\; which\; should\; be\; the\; indices\; of\; students\; in\; the\; team\; you\; form.\; If\; there\; are\; multiple\; answers,\; print\; any\; of\; them.}$

Input

The first line contains two integers $\mathrm{nn and kk (1\le k\le n\le 100)\; —\; the\; number\; of\; students\; and\; the\; size\; of\; the\; team\; you\; have\; to\; form.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,an (1\le ai\le 100),\; where aiai is\; the\; rating\; of ii-th\; student.}$

Output

If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $\mathrm{kk distinct\; integers\; from 11 to nnwhich\; should\; be\; the\; indices\; of\; students\; in\; the\; team\; you\; form.\; All\; the\; ratings\; of\; the\; students\; in\; the\; team\; should\; be\; distinct.\; You\; may\; print\; the\; indices\; in\; any\; order.\; If\; there\; are\; multiple\; answers,\; print\; any\; of\; them.}$

Assume that the students are numbered from $\mathrm{1 to n.}$

Examples

input

Copy

5 3
15 13 15 15 12

output

Copy

YES
1 2 5 

input

Copy

5 4
15 13 15 15 12

output

Copy

NO

input

Copy

4 4
20 10 40 30

output

Copy

YES
1 2 3 4 

Note

All possible answers for the first example:

• {1 2 5}
• {2 3 5}
• {2 4 5}

Note that the order does not matter.

 题意:n个数字，求输出m个不同的数字的下标 不足m个输出NO

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int f[105],pos[105];
int main()
{
    int n,m;
    int cnt=0,d;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&d);
        f[d]++;
        if(f[d]==1)
            {cnt++;
            pos[d]=i;}
    }
    if(cnt>=m)
    {
        printf("YES\n");
        for(int i=1;i<=100&&m;i++)
        {
            if(pos[i])
            {
                printf("%d ",pos[i]);
                m--;
            }
        }
    }
    else
        printf("NO\n");

    return 0;
}

B. Substrings Sort

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given $\mathrm{nn strings.\; Each\; string\; consists\; of\; lowercase\; English\; letters.\; Rearrange\; (reorder)\; the\; given\; strings\; in\; such\; a\; way\; that\; for\; every\; string,\; all\; strings\; that\; are\; placed\; before\; it\; are\; its substrings.}$

String $\mathrm{aa is\; a substring of\; string b if\; it\; is\; possible\; to\; choose\; several consecutive letters\; in b in\; such\; a\; way\; that\; they\; form aa.\; For\; example,\; string\; "for"\; is\; contained\; as\; a substring in\; strings\; "codeforces",\; "for"\; and\; "therefore",\; but\; is\; not\; contained\; as\; a substring in\; strings\; "four",\; "fofo"\; and\; "rof".}$

Input

The first line contains an integer $\mathrm{nn (1\le n\le 100)\; —\; the\; number\; of\; strings.}$

The next $\mathrm{nn lines\; contain\; the\; given\; strings.\; The\; number\; of\; letters\; in\; each\; string\; is\; from 1 to 100,\; inclusive.\; Each\; string\; consists\; of\; lowercase\; English\; letters.}$

Some strings might be equal.

Output

If it is impossible to reorder $\mathrm{nn given\; strings\; in\; required\; order,\; print\; "NO"\; (without\; quotes).}$

Otherwise print "YES" (without quotes) and $\mathrm{nn given\; strings\; in\; required\; order.}$

Examples

input

Copy

5
a
aba
abacaba
ba
aba

output

Copy

YES
a
ba
aba
aba
abacaba

input

Copy

5
a
abacaba
ba
aba
abab

output

Copy

NO

input

Copy

3
qwerty
qwerty
qwerty

output

Copy

YES
qwerty
qwerty
qwerty

Note

In the second example you cannot reorder the strings because the string "abab" is not a substring of the string "abacaba".

 题意:输出n个字符串，重新排列，让前一个字符串成为这个字符串的子串，如果对于所有字符串都满足，输出YES且输出排列顺序，反之输出NO

写个cmp重载，然后使用substr 当然使用find strstr都行，我就记得substr用法emmmm

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
bool cmp(string a,string b)
{
    if(a.size()==b.size())
        return a>b;
    else return a.size()<b.size();
}
int main()
{
    int n;
    scanf("%d",&n);
    string s[105];
    for(int i=0; i<n; i++)
        cin>>s[i];
    sort(s,s+n,cmp);
    int cnt=0;
    for(int i=1; i<n; i++)
    {
        int len=s[i-1].size();
        for(int j=0; j<s[i].size(); j++)
        {
            if(s[i-1]==s[i].substr(j,len))
            {
                cnt++;
                break;
            }
        }
    }
    if(cnt==n-1)
    {
        cout<<"YES"<<endl;
        for(int i=0; i<n; i++)
            cout<<s[i]<<endl;
    }
    else
        cout<<"NO"<<endl;
}

 

C. Equal Sums

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given $\mathrm{kk sequences\; of\; integers.\; The\; length\; of\; the ii-th\; sequence\; equals\; to nini.}$

You have to choose exactly two sequences $\mathrm{ii and jj (i\ne ji\ne j)\; such\; that\; you\; can\; remove\; exactly\; one\; element\; in\; each\; of\; them\; in\; such\; a\; way\; that\; the\; sum\; of\; the\; changed\; sequence ii (its\; length\; will\; be\; equal\; to ni-1ni-1)\; equals\; to\; the\; sum\; of\; the\; changed\; sequence jj (its\; length\; will\; be\; equal\; to nj-1nj-1).}$

Note that it's required to remove exactly one element in each of the two chosen sequences.

Assume that the sum of the empty (of the length equals $\mathrm{00)\; sequence\; is 00.}$

Input

The first line contains an integer $\mathrm{kk (2\le k\le 2\cdot 105)\; —\; the\; number\; of\; sequences.}$

Then $\mathrm{kk pairs\; of\; lines\; follow,\; each\; pair\; containing\; a\; sequence.}$

The first line in the $\mathrm{ii-th\; pair\; contains\; one\; integer nini (1\le ni<2\cdot 105)\; —\; the\; length\; of\; the ii-th\; sequence.\; The\; second\; line\; of\; the ii-th\; pair\; contains\; a\; sequence\; of nini integers ai,1,ai,2,\dots ,ai,niai,1,ai,2,\dots ,ai,ni.}$

The elements of sequences are integer numbers from $-104-104 to 104104.$

The sum of lengths of all given sequences don't exceed $\mathrm{2\cdot 105,\; i.e. n1+n2+\cdots +nk\le 2\cdot 105.}$

Output

If it is impossible to choose two sequences such that they satisfy given conditions, print "NO" (without quotes). Otherwise in the first line print "YES" (without quotes), in the second line — two integers $\mathrm{ii, xx (1\le i\le k,1\le x\le ni1\le i\le k,1\le x\le ni),\; in\; the\; third\; line\; —\; two\; integers jj, yy (1\le j\le k,1\le y\le nj1\le j\le k,1\le y\le nj).\; It\; means\; that\; the\; sum\; of\; the\; elements\; of\; the ii-th\; sequence\; without\; the\; element\; with\; index xx equals\; to\; the\; sum\; of\; the\; elements\; of\; the jj-th\; sequence\; without\; the\; element\; with\; index yy.}$

Two chosen sequences must be distinct, i.e. $\mathrm{i\ne ji\ne j.\; You\; can\; print\; them\; in\; any\; order.}$

If there are multiple possible answers, print any of them.

Examples

input

Copy

2
5
2 3 1 3 2
6
1 1 2 2 2 1

output

Copy

YES
2 6
1 2

input

Copy

3
1
5
5
1 1 1 1 1
2
2 3

output

Copy

NO

input

Copy

4
6
2 2 2 2 2 2
5
2 2 2 2 2
3
2 2 2
5
2 2 2 2 2

output

Copy

YES
2 2
4 1

Note

In the first example there are two sequences $[2,3,1,3,2]and [1,1,2,2,2,1] You\; can\; remove\; the\; second\; element\; from\; the\; first\; sequence\; to\; get [2,1,3,2] and\; you\; can\; remove\; the\; sixth\; element\; from\; the\; second\; sequence\; to\; get [1,1,2,2,2][1,1,2,2,2].\; The\; sums\; of\; the\; both\; resulting\; sequences\; equal\; to 88,\; i.e.\; the\; sums\; are\; equal.$

 题意:给出K个序列，求是否可以删除两个序列的的一个元素，使得两个元素和相等

map套pair map记录每次和，然后pari记录删除的序列及元素

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
map<int,pair<int,int> >mp;
const int maxn = 200000+5;
int a[maxn];
int main()
{
    int n,k,sum;
    scanf("%d",&k);
    for(int i=1; i<=k; i++)
    {
        sum=0;
        scanf("%d",&n);
        for(int j=1; j<=n; j++)
        {
            scanf("%d",&a[j]);
            sum+=a[j];
        }
        for(int j=1;j<=n;j++)
        {
            map<int, pair<int,int> >::iterator it=mp.find(sum-a[j]);
            if(it!=mp.end())
            {
                printf("YES\n");
                printf("%d %d\n%d %d\n",it->second.first,it->second.second,i,j);
                return 0;
            }
        }
        for(int j=1;j<=n;j++)
        {
            map<int, pair<int,int> >::iterator it=mp.find(sum-a[j]);
            if(it==mp.end())
            {
               pair<int,int> p=make_pair(i,j);
                mp[sum-a[j]]=p;
            }
        }
    }
    printf("NO\n");
    return 0;
}

 

D. Points and Powers of Two

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are $\mathrm{nn distinct\; points\; on\; a\; coordinate\; line,\; the\; coordinate\; of ii-th\; point\; equals\; to xi.\; Choose\; a\; subset\; of\; the\; given\; set\; of\; points\; such\; that\; the\; distance\; between\; each\; pair\; of\; points\; in\; a\; subset\; is\; an\; integral\; power\; of\; two.\; It\; is\; necessary\; to\; consider\; each\; pair\; of\; points,\; not\; only\; adjacent.\; Note\; that\; any\; subset\; containing\; one\; element\; satisfies\; the\; condition\; above.\; Among\; all\; these\; subsets,\; choose\; a\; subset\; with\; maximum\; possible\; size.}$

In other words, you have to choose the maximum possible number of points ${\mathrm{xi1,xi2,\dots ,xim such\; that\; for\; each\; pair xijxij, xikxik it\; is\; true\; that |xij-xik|=2d where dd is\; some\; non-negative\; integer\; number\; (not\; necessarily\; the\; same\; for\; each\; pair\; of\; points).}}^{}$

Input

The first line contains one integer $\mathrm{nn (1\le n\le 2\cdot 105)\; —\; the\; number\; of\; points.}$

The second line contains $\mathrm{nn pairwise\; distinct\; integers x1,x2,\dots ,xn (-109\le xi\le 109)\; —\; the\; coordinates\; of\; points.}$

Output

In the first line print $\mathrm{mm —\; the\; maximum\; possible\; number\; of\; points\; in\; a\; subset\; that\; satisfies\; the\; conditions\; described\; above.}$

In the second line print $\mathrm{mm integers\; —\; the\; coordinates\; of\; points\; in\; the\; subset\; you\; have\; chosen.}$

If there are multiple answers, print any of them.

Examples

input

Copy

6
3 5 4 7 10 12

output

Copy

3
7 3 5

input

Copy

5
-1 2 5 8 11

output

Copy

1
8

Note

In the first example the answer is $[7,3,5].\; Note,\; that |7-3|=4=22, |7-5|=2=21 and |3-5|=2=21.\; You\; can\text{'}t\; find\; a\; subset\; having\; more\; points\; satisfying\; the\; required\; property.$

 题意：求一个集合中的任意两个差为2的整数幂次

思路：开始一直没想到集合元素最多三个，后面补题才想到emmm，直接把元素都如set，然后直接判断a[i]-(1<<j)与a[i]+(1<<j)在set出现过没就行

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 200000+5;
set<ll> st;
ll a[maxn];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
    {
        scanf("%lld",&a[i]);
        st.insert(a[i]);
    }
    for(int i=1; i<=n; i++)
    {
        for(int j=0; j<31; j++)
        {
            if(st.find(a[i]+(1LL*1<<j))!=st.end()&&st.find(a[i]-(1LL*1<<j))!=st.end())
            {
                printf("3\n%lld %lld %lld\n",a[i]+(1LL*1<<j),a[i],a[i]-(1LL*1<<j));
                return 0;
            }
        }
    }
    for(int i=1; i<=n; i++)
    {
        for(int j=0; j<31; j++)
        {
            if(st.find(a[i]+(1LL*1<<j))!=st.end())
            {
                printf("2\n%lld %lld\n",a[i],a[i]+1LL*(1<<j));
                return 0;
            }
        }
    }
    printf("1\n%lld\n",a[1]);
}

E. Divisibility by 25

 

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an integer $\mathrm{nn from 11 to 1018 without\; leading\; zeroes.}$

In one move you can swap any two adjacent digits in the given number in such a way that the resulting number will not contain leading zeroes. In other words, after each move the number you have cannot contain any leading zeroes.

What is the minimum number of moves you have to make to obtain a number that is divisible by $\mathrm{25?\; Print -1 if\; it\; is\; impossible\; to\; obtain\; a\; number\; that\; is\; divisible\; by 25.}$

Input

The first line contains an integer $\mathrm{nn (1\le n\le 1018).\; It\; is\; guaranteed\; that\; the\; first\; (left)\; digit\; of\; the\; number nn is\; not\; a\; zero.}$

Output

If it is impossible to obtain a number that is divisible by $\mathrm{25,\; print -1.\; Otherwise\; print\; the\; minimum\; number\; of\; moves\; required\; to\; obtain\; such\; number.}$

Note that you can swap only adjacent digits in the given number.

Examples

input

Copy

5071

output

Copy

4

input

Copy

705

output

Copy

1

input

Copy

1241367

output

Copy

-1

Note

In the first example one of the possible sequences of moves is 5071 $\to \to 5701 \to \to 7501 \to \to 7510 \to \to 7150.$

 题意：给出一个数，求它能否交换(相邻的交换 最后能整除25 

就考虑末尾为00 75 25 50的情况，注意交换后出现前导0的情况。直接模拟就好了  比赛的时候模拟半天写错了，还是刷少了题

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 20;
char a[maxn];
int b[maxn];
int ans=0x3f3f3f3f;
int f[11];
void pd(int x,int y)
{
    int i;
    for(int i=1; i<=strlen(a+1); i++)
        b[i]=a[i]-'0';
    int cnt=0;
    int n=strlen(a+1);

    if(b[n]!=y)
    {
        for( i=n; i>=0; i--)
            if(b[i]==y) break;
        for(; i<n; i++)
        {
            swap(b[i],b[i+1]);
            cnt++;
        }

    }
    if(b[n-1]!=x)
    {
        for(i=n-1; i>=0; i--)
            if(b[i]==x) break;
        for(; i<n-1; i++)
        {
            swap(b[i],b[i+1]);
            cnt++;
        }
    }
    if(b[1]==0)
    {
        for(i=2; i<=n-2; i++)
            if(b[i]!=0) break;
        if(i==n-1) return;
        cnt+=i-1;
    }
    ans=min(ans,cnt);;
}
int main()
{
    scanf("%s",a+1);
    bool fg=0;
    for(int i=1; i<=strlen(a+1); i++)
    {
        f[a[i]-'0']++;
    }
    if(f[0]>=2)
    {
        pd(0,0);
        fg=true;
    }
    if(f[5]>=1&&f[2]>=1)
    {
        pd(2,5);
        fg=true;
    }
    if(f[5]>=1&&f[7]>=1)
    {
        pd(7,5);
        fg=true;
    }
    if(f[5]>=1&&f[0]>=1)
    {
        pd(5,0);
        fg=true;
    }
    if(fg)
        printf("%d\n",ans);
    else
        printf("-1\n");
}

 

 F  还没补emmmmm  告辞

 

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