@Adu Vai 404 attended on "Codeforce Round 1101"

Photo courtesy Idea: 36champ

Solution Idea: 36champ

If the meeting situation is fixed, the answer is the number of people to the left of that position, the number of people to the right of that position. #definine pb push back

We're talking namespace std;

typedef long long longll;

typedef vector (vi);

typedef vector Vvi;

typedef vector I'm sorry.

I don't know. I don't know, pii;

I don't know. I don't know, pll;

Intmain()

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OS base:: sync with stdio (false);

cin.tie (nulrptr);

Cout.tie (nulrptr);

Int t = 1;

cin> > t;

While (t-->0)

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I don't know.

I'm sorry.

vi a (n);

for (int i = 0; i)> a[i];

sort (a.begin(), a.end());

Int cnt = 0;

for (int i = 0; i)#definine pb push back

We're talking namespace std;

typedef long long longll;

typedef vector (vi);

typedef vector Vvi;

typedef vector I'm sorry.

I don't know. I don't know, pii;

I don't know. I don't know, pll;

Intmain()

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OS base:: sync with stdio (false);

cin.tie (nulrptr);

Cout.tie (nulrptr);

Int t = 1;

cin> > t;

While (t-->0)

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I don't know.

I'm sorry.

vll a(n);

for (int i = 0; i)> a[i];

♪ Ans = 1e9, sum = 0;

for (int i = 0; i) #Include #definine pb push back

We're talking namespace std;

typedef long long longll;

typedef vector (vi);

typedef vector Vvi;

typedef vector I'm sorry.

I don't know. I don't know, pii;

I don't know. I don't know, pll;

I'll s, t;

ll eval (string & S, int m)

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ll T = 0, ans = 0;

& for

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If(c = 'I')

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If (T < t)

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T++;

Ans++;

♪ I'm sorry ♪

If (c = 'E')

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If(ans< T*s)ans++;

♪ I'm sorry ♪

I'm sorry, else.

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If(m->0)

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If (T < t)

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T++;

Ans++;

♪ I'm sorry ♪

I'm sorry, else.

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If(ans< T*s)ans++;

♪ I'm sorry ♪

I'm sorry.

♪ I'm sorry ♪

Intmain()

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OS base:: sync with stdio (false);

cin.tie (nulrptr);

Cout.tie (nulrptr);

Int T = 1;

Cin > T;

While

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I don't know.

cin > n > t > s;

I don't know.

cin > S;

Int l = 0, r = 0;

For (charc:S) if(c = 'A') r++;

♪ Ans = 0;

For (int i=l; i<=r; i++) ans = max (ans, eval(S, i));

What's going on?

♪ I'm sorry ♪

$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ million if the first US$$$$$$$$$ million if it acts as funds.$$$$.$$$.$$$.$$.$$$.$$$.00 if the first$$$$$$$$$$.00, what is the change?

$1.$.$.$.00.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$.$. #definine pb push back

We're talking namespace std;

typedef long long longll;

typedef vector (vi);

typedef vector Vvi;

typedef vector I'm sorry.

I don't know. I don't know, pii;

I don't know. I don't know, pll;

I'll s, t;

ll eval (string & S, int m)

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ll T = 0, ans = 0;

& for

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If(c = 'I')

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If (T < t)

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T++;

Ans++;

♪ I'm sorry ♪

If (c = 'E')

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If(ans< T*s)ans++;

♪ I'm sorry ♪

I'm sorry, else.

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If(m->0)

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If (T < t)

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T++;

Ans++;

♪ I'm sorry ♪

I'm sorry, else.

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If(ans< T*s)ans++;

♪ I'm sorry ♪

I'm sorry.

♪ I'm sorry ♪

Intmain()

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OS base:: sync with stdio (false);

cin.tie (nulrptr);

Cout.tie (nulrptr);

Int T = 1;

Cin > T;

While

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I don't know.

cin > n > t > s;

I don't know.

cin > S;

Int l = 0, r = 0;

For (charc:S) if(c = 'A') r++;

While (l < r)

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Int m = (l + r) / 2;

Int x = eval (S, m), y = eval (S, m + 1);

If(x < y)l = m + 1;

Orse r = m;

♪ I'm sorry ♪

Cout < eval(S,l) < "\n";

♪ I'm sorry ♪

If we fix the numbers that were given as controls, what is the best way to assign who is transferred/exclused? Therefore, if $1$$$$$$$$$$$$$$$$$$ million, and $1$$$$$$$ million, we want to sign them a non-empty table. #definine pb push back

We're talking namespace std;

typedef long long longll;

typedef vector (vi);

typedef vector Vvi;

typedef vector I'm sorry.

I don't know. I don't know, pii;

I don't know. I don't know, pll;

Intmain()

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OS base:: sync with stdio (false);

cin.tie (nulrptr);

Cout.tie (nulrptr);

Int T = 1;

Cin > T;

While

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All right, s, t;

cin > n > t > s;

I don't know.

cin > S;

Alll = 0, r = 0, ans = 0;

& for

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If(c = = 'A')

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If (t * s = ans) continue;

Ans++;

If(s > l*s) l++;

r = min (t, r + 1);

♪ I'm sorry ♪

If (c = 'I')

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If (l = t) continue;

Ans++;

l++;

r = min (t, r + 1);

♪ I'm sorry ♪

If (c = 'E')

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If(es = r*s) continue;

Ans++;

If(s > l*s) l++;

♪ I'm sorry ♪

What's going on?

♪ I'm sorry ♪

Photo courtesy Idea: 36champ

Solution Idea: 36champ

If it is affordable because there can't be anything else, we can move the cause in at least $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ million for the first place.

This takes $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ million to move the $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$...$$$$$$$$.$$$$$$$$$$$$$$$$$$$$$$$$$ million to move the first $1$$$$$$$$$$$$$$$ million...$$$$$$

Case 2: $$a k\geq $1$

This takes $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$.$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$...$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$...$$$$$$$$$$$.$.$.$.00 (k-$$$$$$$$$.00-$.00)$$$$$$$

#Include #definine pb push back

We're talking namespace std;

typedef long long longll;

typedef vector (vi);

typedef vector Vvi;

typedef vector I'm sorry.

I don't know. I don't know, pii;

I don't know. I don't know, pll;

I don't know.

vi a (20);

vvi op;

void moveall (int k, int s, int t)

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If(k < = 0) return;

Int u = 6-s-t;

If(a[k-1] = 0)

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Moveall (k-1, s, u);

op.pb({k, s, });

Moveall (k-1, u, t);

♪ I'm sorry ♪

I'm sorry, else.

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Moveall (k-1-a[k-1], s, u);

op.pb({k, s, });

moveall(k-1-a[k-1],u, s);

Moveall (k-1, s, t);

♪ I'm sorry ♪

Intmain()

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OS base:: sync with stdio (false);

cin.tie (nulrptr);

Cout.tie (nulrptr);

Int t;

cin> > t;

While (t-->0)

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I'm sorry.

for (int i = 0; i)> a[i];

Bool ok = true;

for (int i = 0; i) (i) ok = false;

If(!ok)

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What's going on?

I'm sorry.

♪ I'm sorry ♪

What's going on?

Moveall (n, 1, 3);

Cout < < op. size()< "\n";

For(vi p: op) cout < < p[0] < "< p [1] < < " < < p[2] < "\n";

Op.clar();

♪ I'm sorry ♪

Photo courtesy Idea: 36champ

Solution Idea: 36champ

Look at the antti-diagonals (bottom left- > top right). There is a bijection between the snaking arrengement and the situation of $$$$ [1, 2, \dots, ] $$. By looking at ant-diagonals of the grid, we can examine that first ant-diagonal (just the top-leaf cell) only contains the snake of Lingth $$$$$$$$$$$.00.

Three, three, three, three.

Three.

♪ Now that that 2 is to the left of 3. ♪ ♪ Now that that 1 is to the right of 3. This can now be sold together in $$$$$$O (n)2 or in $$$$O (n) with more optimation. ♪ #definine pb push back

We're talking namespace std;

typedef long long longll;

typedef vector (vi);

typedef vector Vvi;

typedef vector I'm sorry.

I don't know. I don't know, pii;

I don't know. I don't know, pll;

I don't know.

#definine max N200005

vll fac (MAX N, 1);

Intmain()

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OS base:: sync with stdio (false);

cin.tie (nulrptr);

Cout.tie (nulrptr);

for(int i=2; i)> t;

While (t-->0)

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Okay, n, k;

cin > n > k;

VviD;vid;

Bool ok = true;

for (int i = 0; i)> s > r > c;

/ Starting Point Check

If(s / 2 + r + c! = n + 1) ok = false;

vector V;

V.pb({r, c});

If(s! = 1)

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I don't know.

cin > S;

For (char C:S)

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If(C = 'R')

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c++;

V.pb({r, c});

♪ I'm sorry ♪

I'm sorry, else.

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r++;

V.pb({r, c});

♪ I'm sorry ♪

/ Symmetry Check

for (int i = 0; i) * This can be quoted another using US$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$0.00$$$$$$$$$$$$$$$$$$$$.00.00 dollars * (a + b) *$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$0.00$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$0.00$$$$$$$$$$$$$$$$$$$$$$0.00$$$$$$$$$$$$$$$$$0.00$$$$$$$$$$ dollars dollars dollars dollars$$$$$$$$$$$$$$$

The \lt b$$$$$ case can be Handled simple. #definine pb push back

We're talking namespace std;

typedef long long longll;

typedef vector (vi);

typedef vector Vvi;

typedef vector I'm sorry.

I don't know. I don't know, pii;

I don't know. I don't know, pll;

/ Faction to find the solution

ll gcd(ll a,ll b,ll & x,ll &y){

If(b = 0){

x = 1; y = 0;

I'm sorry.

♪ I'm sorry ♪

ll x 1, y1;

ll d = gcd(b, a%b, x 1, y1);

x = y1;

y = x1 - y1* (a/b);

I don't know.

♪ I'm sorry ♪

Intmain()

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OS base:: sync with stdio (false);

cin.tie (nulrptr);

Cout.tie (nulrptr);

Int t = 1;

cin> > t;

While (t-->0)

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Okay, n, a, b, k;

cin > n > a > b > k;

ll g = gcd(a, b);

If(k %g! = 0)

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I'm sorry.

♪ I'm sorry ♪

a = g; b = g; k = g;

If(a = = 1 & b = 1)

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What's going on?

I'm sorry.

♪ I'm sorry ♪

/ / STEP 1

♪ Ans = 0;

If(k %a = 0)

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is++; n--

ll x = k/a;

While(n > 0)

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If(k-(x*b) < 0 || (k-x*b) %a!=0) break;

x = (k-x*b) / a;

Ans++;

No, no, no.

♪ I'm sorry ♪

/ / STEP 2

ll v = 0;

If(k % (a + b) = 0) v = 1e9;

I'm sorry, else.

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If (a < b) swap (a, b);

// SOLVE(a + b) x + b *a^v y = k

ll m = b;

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ll x, y;

gcd (a + b, m, x, y);

If(x< 0){x + = m; y -=a + b;}

x* = k; y * = k;

ll d = x/m;

x - = d * m; y + = d * (a + b);

/cout < x < " "< y < "\n";

/getchar();

If (m / b * y + x < 0) break;

// Security Check

Bool ok = true;

ll x1 = x, x2 = m / b * y + x;

k / m) break;

m* = a;

♪ I'm sorry ♪

/ Cout < "! "< v < " \n";

cout <

♪ I'm sorry ♪