Educational Codeforces Round 49

A. Palindromic Twist

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string $\mathrm{ss consisting\; of nn lowercase\; Latin\; letters. nn is\; even.}$

For each position $\mathrm{ii (1\le i\le n1\le i\le n)\; in\; string ss you\; are\; required\; to\; change\; the\; letter\; on\; this\; position\; either\; to\; the\; previous\; letter\; in\; alphabetic\; order\; or\; to\; the\; next\; one\; (letters\; \text{'}a\text{'}\; and\; \text{'}z\text{'}\; have\; only\; one\; of\; these\; options).\; Letter\; in\; every\; position\; must\; be\; changed exactly\; once.}$

For example, letter 'p' should be changed either to 'o' or to 'q', letter 'a' should be changed to 'b' and letter 'z' should be changed to 'y'.

That way string "codeforces", for example, can be changed to "dpedepqbft" ('c' $\to \to \text{'}d\text{'},\; \text{'}o\text{'} \to \to \text{'}p\text{'},\; \text{'}d\text{'} \to \to \text{'}e\text{'},\; \text{'}e\text{'} \to \to \text{'}d\text{'},\; \text{'}f\text{'} \to \to \text{'}e\text{'},\; \text{'}o\text{'} \to \to \text{'}p\text{'},\; \text{'}r\text{'} \to \to \text{'}q\text{'},\; \text{'}c\text{'} \to \to \text{'}b\text{'},\; \text{'}e\text{'} \to \to \text{'}f\text{'},\; \text{'}s\text{'} \to \to \text{'}t\text{'}).$

String $\mathrm{ss is\; called\; a\; palindrome\; if\; it\; reads\; the\; same\; from\; left\; to\; right\; and\; from\; right\; to\; left.\; For\; example,\; strings\; "abba"\; and\; "zz"\; are\; palindromes\; and\; strings\; "abca"\; and\; "zy"\; are\; not.}$

Your goal is to check if it's possible to make string $\mathrm{ss a\; palindrome\; by\; applying\; the\; aforementioned\; changes\; to\; every\; position.\; Print\; "YES"\; if\; string ss can\; be\; transformed\; to\; a\; palindrome\; and\; "NO"\; otherwise.}$

Each testcase contains several strings, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer $\mathrm{TT (1\le T\le 501\le T\le 50)\; —\; the\; number\; of\; strings\; in\; a\; testcase.}$

Then $\mathrm{2T2T lines\; follow\; —\; lines (2i-1)(2i-1) and 2i2i of\; them\; describe\; the ii-th\; string.\; The\; first\; line\; of\; the\; pair\; contains\; a\; single\; integer nn (2\le n\le 1002\le n\le 100, nn is\; even)\; —\; the\; length\; of\; the\; corresponding\; string.\; The\; second\; line\; of\; the\; pair\; contains\; a\; string ss,\; consisting\; of nnlowercase\; Latin\; letters.}$

Output

Print $\mathrm{TT lines.\; The ii-th\; line\; should\; contain\; the\; answer\; to\; the ii-th\; string\; of\; the\; input.\; Print\; "YES"\; if\; it\text{'}s\; possible\; to\; make\; the ii-th\; string\; a\; palindrome\; by\; applying\; the\; aforementioned\; changes\; to\; every\; position.\; Print\; "NO"\; otherwise.}$

Example

input

Copy

5
6
abccba
2
cf
4
adfa
8
abaazaba
2
ml

output

Copy

YES
NO
YES
NO
NO

Note

The first string of the example can be changed to "bcbbcb", two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.

The second string can be changed to "be", "bg", "de", "dg", but none of these resulting strings are palindromes.

The third string can be changed to "beeb" which is a palindrome.

The fifth string can be changed to "lk", "lm", "nk", "nm", but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can't obtain strings "ll" or "mm".

反正每个字母都要改变，那么相对的字母要不相差为二，要不为零．

#include <bits/stdc++.h>
using namespace std;
int main(){
    int n;
    cin>>n;
    while(n--){
        int x;
        char s[101];
        cin>>x>>s;
        int l = 0,r = x-1;
        bool prime = true;
        for(int i=1;i<=x/2;i++){
            int xn = s[l]-'a';
            int y = s[r]-'a';
            if(abs(xn-y)!=2&&abs(xn-y)!=0){
                prime = false;
                break;
            }else{
                l++,r--;
            }
        }
        if(prime){
            cout<<"YES"<<endl;
        }else{
            cout<<"NO"<<endl;
        }
    }
    return 0;
}

B. Numbers on the Chessboard

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a chessboard of size $\mathrm{n\times nn\times n.\; It\; is\; filled\; with\; numbers\; from 11 to n2n2 in\; the\; following\; way:\; the\; first \lceil n22\rceil \lceil n22\rceil numbers\; from 11 to \lceil n22\rceil \lceil n22\rceil are\; written\; in\; the\; cells\; with\; even\; sum\; of\; coordinates\; from\; left\; to\; right\; from\; top\; to\; bottom.\; The\; rest n2-\lceil n22\rceil n2-\lceil n22\rceil numbers\; from \lceil n22\rceil +1\lceil n22\rceil +1 to n2n2are\; written\; in\; the\; cells\; with\; odd\; sum\; of\; coordinates\; from\; left\; to\; right\; from\; top\; to\; bottom.\; The\; operation \lceil xy\rceil \lceil xy\rceil means\; division xx by yy rounded\; up.}$

For example, the left board on the following picture is the chessboard which is given for $\mathrm{n=4n=4 and\; the\; right\; board\; is\; the\; chessboard\; which\; is\; given\; for n=5n=5.}$

You are given $\mathrm{qq queries.\; The ii-th\; query\; is\; described\; as\; a\; pair xi,yixi,yi.\; The\; answer\; to\; the ii-th\; query\; is\; the\; number\; written\; in\; the\; cell xi,yixi,yi (xixiis\; the\; row, yiyi is\; the\; column).\; Rows\; and\; columns\; are\; numbered\; from 11 to nn.}$

Input

The first line contains two integers $\mathrm{nn and qq (1\le n\le 1091\le n\le 109, 1\le q\le 1051\le q\le 105)\; —\; the\; size\; of\; the\; board\; and\; the\; number\; of\; queries.}$

The next $\mathrm{qq lines\; contain\; two\; integers\; each.\; The ii-th\; line\; contains\; two\; integers xi,yixi,yi (1\le xi,yi\le n1\le xi,yi\le n)\; —\; description\; of\; the ii-th\; query.}$

Output

For each query from $\mathrm{11 to qq print\; the\; answer\; to\; this\; query.\; The\; answer\; to\; the ii-th\; query\; is\; the\; number\; written\; in\; the\; cell xi,yixi,yi (xixi is\; the\; row, yiyi is\; the\; column).\; Rows\; and\; columns\; are\; numbered\; from 11 to nn.\; Queries\; are\; numbered\; from 11 to qq in\; order\; of\; the\; input.}$

Examples

input

Copy

4 5
1 1
4 4
4 3
3 2
2 4

output

Copy

1
8
16
13
4

input

Copy

5 4
2 1
4 2
3 3
3 4

output

Copy

16
9
7
20

Note

Answers to the queries from examples are on the board in the picture from the problem statement.

看好规律就可以，但是要用ｌｏｎｇｌｏｎｇ，为此哇了一发．

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll n,m;
int main(){
    cin>>n>>m;
    while(m--){
        ll x,y;
        cin>>x>>y;
        ll ans = (x-1)*n+y;
        ll begin = (n*n+1)/2;
        if(n%2){
            if(x%2){
                if(ans%2){
                    ll xn = (x-1)/2*n;
                    y = (y+1)/2;
                    cout<<xn+y<<endl;
                }else{
                    ll xn = (x-1)/2*n;
                    y = y/2;
                    cout<<begin+xn+y<<endl;
                }
            }else{
                if(ans%2==0){
                    ll xn = (x)/2*n-(n+1)/2;
                    y = (y+1)/2;
                    cout<<begin+xn+y<<endl;
                }else{
                    ll xn = (x)/2*n-n/2;
                    y = y/2;
                    cout<<xn+y<<endl;
                }
            }
        }else{
            if(x%2){
                if(ans%2){
                    ll xn = (x-1)/2*n;
                    y = (y+1)/2;
                    cout<<xn+y<<endl;
                }else{
                    ll xn = (x-1)/2*n;
                    y = y/2;
                    cout<<begin+xn+y<<endl;
                }
            }else{
                if(ans%2){
                    ll xn = (x)/2*n-n/2;
                    y = (y+1)/2;
                    cout<<begin+xn+y<<endl;
                }else{
                    ll xn = (x)/2*n-n/2;
                    y = y/2;
                    cout<<xn+y<<endl;
                }
            }
        }
    }
    return 0;
}

C. Minimum Value Rectangle

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have $\mathrm{nn sticks\; of\; the\; given\; lengths.}$

Your task is to choose exactly four of them in such a way that they can form a rectangle. No sticks can be cut to pieces, each side of the rectangle must be formed by a single stick. No stick can be chosen multiple times. It is guaranteed that it is always possible to choose such sticks.

Let $\mathrm{SS be\; the\; area\; of\; the\; rectangle\; and PP be\; the\; perimeter\; of\; the\; rectangle.}$

The chosen rectangle should have the value $\frac{{\mathrm{P2SP2S minimal\; possible.\; The\; value\; is\; taken\; without\; any\; rounding.}}^{}}{}$

If there are multiple answers, print any of them.

Each testcase contains several lists of sticks, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer $\mathrm{TT (T\ge 1T\ge 1)\; —\; the\; number\; of\; lists\; of\; sticks\; in\; the\; testcase.}$

Then $\mathrm{2T2T lines\; follow\; —\; lines (2i-1)(2i-1) and 2i2i of\; them\; describe\; the ii-th\; list.\; The\; first\; line\; of\; the\; pair\; contains\; a\; single\; integer nn (4\le n\le 1064\le n\le 106)\; —\; the\; number\; of\; sticks\; in\; the ii-th\; list.\; The\; second\; line\; of\; the\; pair\; contains nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le aj\le 1041\le aj\le 104)\; —\; lengths\; of\; the\; sticks\; in\; the ii-th\; list.}$

It is guaranteed that for each list there exists a way to choose four sticks so that they form a rectangle.

The total number of sticks in all $\mathrm{TT lists\; doesn\text{'}t\; exceed 106106 in\; each\; testcase.}$

Output

Print $\mathrm{TT lines.\; The ii-th\; line\; should\; contain\; the\; answer\; to\; the ii-th\; list\; of\; the\; input.\; That\; is\; the\; lengths\; of\; the\; four\; sticks\; you\; choose\; from\; the ii-th\; list,\; so\; that\; they\; form\; a\; rectangle\; and\; the\; value P2SP2S of\; this\; rectangle\; is\; minimal\; possible.\; You\; can\; print\; these\; four\; lengths\; in\; arbitrary\; order.}$

If there are multiple answers, print any of them.

Example

input

Copy

3
4
7 2 2 7
8
2 8 1 4 8 2 1 5
5
5 5 5 5 5

output

Copy

2 7 7 2
2 2 1 1
5 5 5 5

Note

There is only one way to choose four sticks in the first list, they form a rectangle with sides $\mathrm{22 and 77,\; its\; area\; is 2\cdot 7=142\cdot 7=14,\; perimeter\; is 2(2+7)=182(2+7)=18. 18214\approx 23.14318214\approx 23.143.}$

The second list contains subsets of four sticks that can form rectangles with sides $(1,2)(1,2), (2,8)(2,8) and (1,8)(1,8).\; Their\; values\; are 622=18622=18, 20216=2520216=25 and 1828=40.51828=40.5,\; respectively.\; The\; minimal\; one\; of\; them\; is\; the\; rectangle (1,2)(1,2).$

You can choose any four of the $\mathrm{55 given\; sticks\; from\; the\; third\; list,\; they\; will\; form\; a\; square\; with\; side 55,\; which\; is\; still\; a\; rectangle\; with\; sides (5,5)(5,5).}$

$\mathrm{这题难点，要是Ｐ＾２／Ｓ化简}$

$\mathrm{最后可以化成x/y+y/x最小，求这个最小，那就是x和y越接近越好．}$

$\mathrm{那么x/y越接近１，越好，所以如果一个为想x/y\; 一个为xx/yy，如果后者大于前者，那麽就取后者，当然}$

$\mathrm{前提是已经排好序．}$

#include <bits/stdc++.h>
#define ll long long
#define N 1000006
using namespace std;

ll a[N], cnt = 0, b[N];
int main() {
    int t, n;
    cin >> t;
    while(t--) {
        cin>>n;
        cnt = 0;
        for(int i = 1; i <= n; i ++) scanf("%lld", &a[i]);
        sort(a+1,a+1+n);
        for(int i = 1; i < n; i ++) {
            if(a[i] == a[i+1]) {
                b[cnt++] = a[i];
                i++;
            }
        }
        ll ans1 = b[0], ans2 = b[1];

        for(int i = 2; i < cnt; i ++) {
            if(ans1*b[i] < ans2*b[i-1]) {
                ans1 = b[i-1];
                ans2 = b[i];
            }
        }
        cout<<ans1<<" "<<ans1<<" "<<ans2<<" "<<ans2<<endl;
    }
    return 0;
}

D. Mouse Hunt

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Medicine faculty of Berland State University has just finished their admission campaign. As usual, about $\mathrm{80\%80\% of\; applicants\; are\; girls\; and\; majority\; of\; them\; are\; going\; to\; live\; in\; the\; university\; dormitory\; for\; the\; next 44 (hopefully)\; years.}$

The dormitory consists of $\mathrm{nn rooms\; and\; a\; single\; mouse!\; Girls\; decided\; to\; set\; mouse\; traps\; in\; some\; rooms\; to\; get\; rid\; of\; the\; horrible\; monster.\; Setting\; a\; trap\; in\; room\; number ii costs cici burles.\; Rooms\; are\; numbered\; from 11 to nn.}$

Mouse doesn't sit in place all the time, it constantly runs. If it is in room $\mathrm{ii in\; second tt then\; it\; will\; run\; to\; room aiai in\; second t+1t+1 without\; visiting\; any\; other\; rooms\; inbetween\; (i=aii=ai means\; that\; mouse\; won\text{'}t\; leave\; room ii).\; It\text{'}s\; second 00 in\; the\; start.\; If\; the\; mouse\; is\; in\; some\; room\; with\; a\; mouse\; trap\; in\; it,\; then\; the\; mouse\; get\; caught\; into\; this\; trap.}$

That would have been so easy if the girls actually knew where the mouse at. Unfortunately, that's not the case, mouse can be in any room from $\mathrm{11 to nn at\; second 00.}$

What it the minimal total amount of burles girls can spend to set the traps in order to guarantee that the mouse will eventually be caught no matter the room it started from?

Input

The first line contains as single integers $\mathrm{nn (1\le n\le 2\cdot 1051\le n\le 2\cdot 105)\; —\; the\; number\; of\; rooms\; in\; the\; dormitory.}$

The second line contains $\mathrm{nn integers c1,c2,\dots ,cnc1,c2,\dots ,cn (1\le ci\le 1041\le ci\le 104)\; — cici is\; the\; cost\; of\; setting\; the\; trap\; in\; room\; number ii.}$

The third line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le n1\le ai\le n)\; — aiai is\; the\; room\; the\; mouse\; will\; run\; to\; the\; next\; second\; after\; being\; in\; room ii.}$

Output

Print a single integer — the minimal total amount of burles girls can spend to set the traps in order to guarantee that the mouse will eventually be caught no matter the room it started from.

Examples

input

Copy

5
1 2 3 2 10
1 3 4 3 3

output

Copy

3

input

Copy

4
1 10 2 10
2 4 2 2

output

Copy

10

input

Copy

7
1 1 1 1 1 1 1
2 2 2 3 6 7 6

output

Copy

2

Note

In the first example it is enough to set mouse trap in rooms $\mathrm{11 and 44.\; If\; mouse\; starts\; in\; room 11 then\; it\; gets\; caught\; immideately.\; If\; mouse\; starts\; in\; any\; other\; room\; then\; it\; eventually\; comes\; to\; room 44.}$

In the second example it is enough to set mouse trap in room $\mathrm{22.\; If\; mouse\; starts\; in\; room 22 then\; it\; gets\; caught\; immideately.\; If\; mouse\; starts\; in\; any\; other\; room\; then\; it\; runs\; to\; room 22 in\; second 11.}$

Here are the paths of the mouse from different starts from the third example:

• $\mathrm{1\to 2\to 2\to \dots 1\to 2\to 2\to \dots ;}$
• $\mathrm{2\to 2\to \dots 2\to 2\to \dots ;}$
• $\mathrm{3\to 2\to 2\to \dots 3\to 2\to 2\to \dots ;}$
• $\mathrm{4\to 3\to 2\to 2\to \dots 4\to 3\to 2\to 2\to \dots ;}$
• $\mathrm{5\to 6\to 7\to 6\to \dots 5\to 6\to 7\to 6\to \dots ;}$
• $\mathrm{6\to 7\to 6\to \dots 6\to 7\to 6\to \dots ;}$
• $\mathrm{7\to 6\to 7\to \dots 7\to 6\to 7\to \dots ;}$

So it's enough to set traps in rooms $\mathrm{22 and 66.}$

这题感觉找到环就行了．

#include <bits/stdc++.h>
#define N 200005
#define ll long long int
using namespace std;
ll a[N],b[N];
bool vis[N],val[N];
std::vector<ll> v[N];
ll cnt = 0;
ll n;

void dfs(ll x){
    vis[x] = 1;
    for(ll i=0;i<v[x].size();i++){
        ll y = v[x][i];
        if(vis[y]==0){
            dfs(y);
        }
    }
}

void solve(ll x){
    val[x] = 1;
    ll y = b[x];
    if(val[y]){
        ll ans = a[x];
        ll xn = y;
        while(xn!=x){
            ans = min(ans,a[xn]);
            xn = b[xn];
        }
        cnt += ans;
    }else{
        solve(y);
    }
}

int main(){
    cin>>n;
    for(ll i=1;i<=n;i++){
        cin>>a[i];
    }
    for(ll i=1;i<=n;i++){
        cin>>b[i];
        v[i].push_back(b[i]);
        v[b[i]].push_back(i);
    }
    for(ll i = 1;i<=n;i++){
        if(vis[i]==0){
            dfs(i);
            solve(i);
        }
    }
    cout<<cnt<<endl;
    return 0;
}