Codeforces Round #492 (Div. 2) [Thanks, uDebug!]

A. Hit the Lottery

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Allen has a LOT of money. He has $\mathrm{nn dollars\; in\; the\; bank.\; For\; security\; reasons,\; he\; wants\; to\; withdraw\; it\; in\; cash\; (we\; will\; not\; disclose\; the\; reasons\; here).\; The\; denominations\; for\; dollar\; bills\; are 11, 55, 1010, 2020, 100100.\; What\; is\; the\; minimum\; number\; of\; bills\; Allen\; could\; receive\; after\; withdrawing\; his\; entire\; balance?}$

Input

The first and only line of input contains a single integer $\mathrm{nn (1\le n\le 1091\le n\le 109).}$

Output

Output the minimum number of bills that Allen could receive.

Examples

input

Copy

125

output

Copy

3

input

Copy

43

output

Copy

5

input

Copy

1000000000

output

Copy

10000000

Note

In the first sample case, Allen can withdraw this with a $\mathrm{100100 dollar\; bill,\; a 2020 dollar\; bill,\; and\; a 55 dollar\; bill.\; There\; is\; no\; way\; for\; Allen\; to\; receive 125125 dollars\; in\; one\; or\; two\; bills.}$

In the second sample case, Allen can withdraw two $\mathrm{2020 dollar\; bills\; and\; three 11 dollar\; bills.}$

In the third sample case, Allen can withdraw $\mathrm{100000000100000000 (ten\; million!) 100100 dollar\; bills.}$

A题:题意:n=a1*100+a2*20+a3*10+a4*5+a5;求 a1+a2+a3+a4+a5的最小值

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    ll n;cin>>n;
    ll num1=0,num2=0,num3=0,num4=0,num5=0;
    num1=n/100;n%=100;
    num2=n/20;n%=20;
    num3=n/10;n%=10;
    num4=n/5;n%=5;
    num5=n;
    cout<<num1+num2+num3+num4+num5<<endl;
    return 0;
} 

View Code

B. World Cup

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Allen wants to enter a fan zone that occupies a round square and has $\mathrm{nn entrances.}$

There already is a queue of ${\mathrm{aiai people\; in\; front\; of\; the ii-th\; entrance.\; Each\; entrance\; allows\; one\; person\; from\; its\; queue\; to\; enter\; the\; fan\; zone\; in\; one\; minute.}}^{}$

Allen uses the following strategy to enter the fan zone:

• Initially he stands in the end of the queue in front of the first entrance.
• Each minute, if he is not allowed into the fan zone during the minute (meaning he is not the first in the queue), he leaves the current queue and stands in the end of the queue of the next entrance (or the first entrance if he leaves the last entrance).

Determine the entrance through which Allen will finally enter the fan zone.

Input

The first line contains a single integer $\mathrm{nn (2\le n\le 1052\le n\le 105) —\; the\; number\; of\; entrances.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (0\le ai\le 1090\le ai\le 109) —\; the\; number\; of\; people\; in\; queues.\; These\; numbers\; do\; not\; include\; Allen.}$

Output

Print a single integer — the number of entrance that Allen will use.

Examples

input

Copy

4
2 3 2 0

output

Copy

3

input

Copy

2
10 10

output

Copy

1

input

Copy

6
5 2 6 5 7 4

output

Copy

6

Note

In the first example the number of people (not including Allen) changes as follows: $[2,3,2,0]\to [1,2,1,0]\to [0,1,0,0][2,3,2,0]\to [1,2,1,0]\to [0,1,0,0].\; The\; number\; in\; bold\; is\; the\; queue\; Alles\; stands\; in.\; We\; see\; that\; he\; will\; enter\; the\; fan\; zone\; through\; the\; third\; entrance.$

In the second example the number of people (not including Allen) changes as follows:$[10,10]\to [9,9]\to [8,8]\to [7,7]\to [6,6]\to [5,5]\to [4,4]\to [3,3]\to [2,2]\to [1,1]\to [0,0][10,10]\to [9,9]\to [8,8]\to [7,7]\to [6,6]\to [5,5]\to [4,4]\to [3,3]\to [2,2]\to [1,1]\to [0,0].$

In the third example the number of people (not including Allen) changes as follows:$[5,2,6,5,7,4]\to [4,1,5,4,6,3]\to [3,0,4,3,5,2]\to [2,0,3,2,4,1]\to [1,0,2,1,3,0]\to [0,0,1,0,2,0]$

B题:题意 >>加全局变量lazy标记即可>>
收获:即1-n变成0-n-1>>更方便的应用了t%n

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int a[N];
int main(){
    std::ios::sync_with_stdio(false);
    int n;cin>>n;int min_=1e9+10;
    for(int i=0;i<n;i++) cin>>a[i],min_=min(min_,a[i]);
    int lazy=min_;
    int t=min_;t%=n;
    while(1){
        if(a[t]-lazy<=0){ cout<<t+1<<endl;return 0;}
        else{lazy++;}
        t++;t%=n;
    }
    return 0;
}

View Code

C. Tesla

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Allen dreams of one day owning a enormous fleet of electric cars, the car of the future! He knows that this will give him a big status boost. As Allen is planning out all of the different types of cars he will own and how he will arrange them, he realizes that he has a problem.

Allen's future parking lot can be represented as a rectangle with $\mathrm{44 rows\; and nn (n\le 50n\le 50)\; columns\; of\; rectangular\; spaces,\; each\; of\; which\; can\; contain\; at\; most\; one\; car\; at\; any\; time.\; He\; imagines\; having kk (k\le 2nk\le 2n)\; cars\; in\; the\; grid,\; and\; all\; the\; cars\; are\; initially\; in\; the\; second\; and\; third\; rows.\; Each\; of\; the\; cars\; also\; has\; a\; different\; designated\; parking\; space\; in\; the\; first\; or\; fourth\; row.\; Allen\; has\; to\; put\; the\; cars\; into\; corresponding\; parking\; places.}$

Illustration to the first example.

However, since Allen would never entrust his cars to anyone else, only one car can be moved at a time. He can drive a car from a space in any of the four cardinal directions to a neighboring empty space. Furthermore, Allen can only move one of his cars into a space on the first or fourth rows if it is the car's designated parking space.

Allen knows he will be a very busy man, and will only have time to move cars at most $\mathrm{2000020000 times\; before\; he\; realizes\; that\; moving\; cars\; is\; not\; worth\; his\; time.\; Help\; Allen\; determine\; if\; he\; should\; bother\; parking\; his\; cars\; or\; leave\; it\; to\; someone\; less\; important.}$

Input

The first line of the input contains two space-separated integers $\mathrm{nn and kk (1\le n\le 501\le n\le 50, 1\le k\le 2n1\le k\le 2n),\; representing\; the\; number\; of\; columns\; and\; the\; number\; of\; cars,\; respectively.}$

The next four lines will contain $\mathrm{nn integers\; each\; between 00 and kk inclusive,\; representing\; the\; initial\; state\; of\; the\; parking\; lot.\; The\; rows\; are\; numbered 11 to 44 from\; top\; to\; bottom\; and\; the\; columns\; are\; numbered 11 to nn from\; left\; to\; right.}$

In the first and last line, an integer $\mathrm{1\le x\le k1\le x\le k represents\; a\; parking\; spot\; assigned\; to\; car xx (you\; can\; only\; move\; this\; car\; to\; this\; place),\; while\; the\; integer 00 represents\; a\; empty\; space\; (you\; can\text{'}t\; move\; any\; car\; to\; this\; place).}$

In the second and third line, an integer $\mathrm{1\le x\le k1\le x\le k represents\; initial\; position\; of\; car xx,\; while\; the\; integer 00 represents\; an\; empty\; space\; (you\; can\; move\; any\; car\; to\; this\; place).}$

Each $\mathrm{xx between 11 and kk appears\; exactly\; once\; in\; the\; second\; and\; third\; line,\; and\; exactly\; once\; in\; the\; first\; and\; fourth\; line.}$

Output

If there is a sequence of moves that brings all of the cars to their parking spaces, with at most $\mathrm{2000020000 car\; moves,\; then\; print mm,\; the\; number\; of\; moves,\; on\; the\; first\; line.\; On\; the\; following mm lines,\; print\; the\; moves\; (one\; move\; per\; line)\; in\; the\; format ii rr cc,\; which\; corresponds\; to\; Allen\; moving\; car ii to\; the\; neighboring\; space\; at\; row rr and\; column cc.}$

If it is not possible for Allen to move all the cars to the correct spaces with at most $\mathrm{2000020000 car\; moves,\; print\; a\; single\; line\; with\; the\; integer -1-1.}$

Examples

input

Copy

4 5
1 2 0 4
1 2 0 4
5 0 0 3
0 5 0 3

output

Copy

6
1 1 1
2 1 2
4 1 4
3 4 4
5 3 2
5 4 2

input

Copy

1 2
1
2
1
2

output

Copy

-1

input

Copy

1 2
1
1
2
2

output

Copy

2
1 1 1
2 4 1

Note

In the first sample test case, all cars are in front of their spots except car $\mathrm{55,\; which\; is\; in\; front\; of\; the\; parking\; spot\; adjacent.\; The\; example\; shows\; the\; shortest\; possible\; sequence\; of\; moves,\; but\; any\; sequence\; of\; length\; at\; most 2000020000 will\; be\; accepted.}$

In the second sample test case, there is only one column, and the cars are in the wrong order, so no cars can move and the task is impossible.

c题:怎末说呢>>说这个题难??还是不难??总之就是说这些题的思路都不难就看你是否想不想的到>>如果想不到的话试图更暴力一点
暴力呀暴力>>暴力也会出现相通的地方的>>>哎 >>赛后补提>>也不想补>>连续两天晚上cf+球赛,两个四点睡觉>>整个人都傻了?
对最坏情况的判断>>都失误>>难受>>
把2.3行看成一个圆>>每次都旋转一格>>然后check一遍>>看能不能插进去>>

#include<bits/stdc++.h>
using namespace std;
using ll = long long int;
using ull = unsigned long long int;
using dd = double;
using ldd = long double;
#define debug(x) cout << #x << " : " << x << '\n'
#define all(x) x.begin(), x.end()
int arr[4][100];int n, k;
queue<pair<pair<int, int>, int>> ans;
bool check()
{
    bool fuh = 0;
    for (int i = 0; i < n; ++i){
        if (arr[0][i] == arr[1][i] && arr[0][i] != 0)
        {
            fuh = 1;
            ans.push({{arr[0][i], 1}, i+1});
            arr[1][i] = 0;
        }
        if (arr[2][i] == arr[3][i] && arr[2][i] != 0)
        {
            fuh = 1;
            ans.push({{arr[2][i], 4}, i+1});
            arr[2][i]=0;
        }
    }
    return fuh;
}
int x, y;
bool circle()
{
    bool ch = 0;
    for (int i = 0; i < 2 * n-1; ++i)
    {
        if (x == 1)
        {
            if (y < n - 1)
            {
                if (arr[x][y+1] == 0)
                {
                    ++y;
                    continue;
                }
                ans.push({{arr[x][y+1], x+1}, y+1});
                arr[x][y] = arr[x][y+1];
                arr[x][y+1]=0;
                ++y;
                ch = 1;
            }
            else
            {
                if (arr[x+1][y] == 0)
                {
                    x++;
                    continue;
                }
                ans.push({{arr[x+1][y], x+1}, y+1});
                arr[x][y] = arr[x+1][y];
                arr[x+1][y] = 0;
                ++x;
                ch = 1;
            }
        }
        else
        {
            if (y > 0)
            {
                if (arr[x][y-1] == 0)
                {
                    --y;
                    continue;
                }
                ans.push({{arr[x][y-1], x+1}, y+1});
                arr[x][y] = arr[x][y-1];
                arr[x][y-1] = 0;
                --y;
                ch = 1;
            }
            else
            {
                if (arr[x-1][y] == 0)
                {
                    --x;
                    continue;
                }
                ans.push({{arr[x-1][y], x+1}, y+1});
                arr[x][y] = arr[x-1][y];
                arr[x-1][y]=0;
                --x;
                ch = 1;
            }
        }
    }
    return ch;
}
int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);
    cin >> n >> k;
    for (int i = 0; i < 4; ++i)
        for (int j = 0; j < n; ++j)
            cin >> arr[i][j];
    if (k == 2 * n)
    {
        if (!check())
        {
            cout << "-1";
            return 0;
        }
    }
    x = y = 0;
    for (int i = 0; i < n; ++i)
    {
        if (arr[1][i] == 0)
        {
            x = 1;
            y = i;
            break;
        }
        if (arr[2][i] == 0)
        {
            x = 2;
            y = i;
            break;
        }
    }
    for (int i = 0; i < 2 * n; ++i)
    {
        check();
        if (circle())
        {
            continue;
        }
        break;
    }
    cout << ans.size() << '\n';
    while (ans.size())
    {
        cout << ans.front().first.first << " " << ans.front().first.second << " " << ans.front().second << '\n';
        ans.pop();
    }
}

View Code

D. Suit and Tie

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Allen is hosting a formal dinner party. $\mathrm{2n2n people\; come\; to\; the\; event\; in nn pairs\; (couples).\; After\; a\; night\; of\; fun,\; Allen\; wants\; to\; line\; everyone\; up\; for\; a\; final\; picture.\; The 2n2n people\; line\; up,\; but\; Allen\; doesn\text{'}t\; like\; the\; ordering.\; Allen\; prefers\; if\; each\; pair\; occupies\; adjacent\; positions\; in\; the\; line,\; as\; this\; makes\; the\; picture\; more\; aesthetic.}$

Help Allen find the minimum number of swaps of adjacent positions he must perform to make it so that each couple occupies adjacent positions in the line.

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1001\le n\le 100),\; the\; number\; of\; pairs\; of\; people.}$

The second line contains $\mathrm{2n2n integers a1,a2,\dots ,a2na1,a2,\dots ,a2n.\; For\; each ii with 1\le i\le n1\le i\le n, ii appears\; exactly\; twice.\; If aj=ak=iaj=ak=i,\; that\; means\; that\; the jj-th\; and kk-th\; people\; in\; the\; line\; form\; a\; couple.}$

Output

Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions.

Examples

input

Copy

4
1 1 2 3 3 2 4 4

output

Copy

2

input

Copy

3
1 1 2 2 3 3

output

Copy

0

input

Copy

3
3 1 2 3 1 2

output

Copy

3

Note

In the first sample case, we can transform $\mathrm{11233244\to 11232344\to 1122334411233244\to 11232344\to 11223344 in\; two\; steps.\; Note\; that\; the\; sequence 11233244\to 11323244\to 1133224411233244\to 11323244\to 11332244 also\; works\; in\; the\; same\; number\; of\; steps.}$

The second sample case already satisfies the constraints; therefore we need $\mathrm{00 swaps.}$

D题:贪心>>每次都先把队首的那个元素找到匹配的元素>>统计下来就是答案>>
凭直觉>>过的

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,tmp;vector<int>a;
int main(){
    std::ios::sync_with_stdio(false);
    cin>>n;for(int i=1;i<=2*n;i++){cin>>tmp;a.push_back(tmp);}
    ll ans=0;
    while(a.size()>0){
        int x=a[0];int pos;
        for(int i=1;i<a.size();i++){
            if(a[i]==x){ans+=(i-1);pos=i;break;}
        }
        auto it=a.begin();a.erase(it+pos);
        it=a.begin();a.erase(it);
    }
    cout<<ans<<endl;
    return 0;
}

View Code

  

E. Leaving the Bar

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

For a vector $\stackrel{}{}$

Allen had a bit too much to drink at the bar, which is at the origin. There are $\mathrm{nn vectors \to v1,\to v2,\cdots ,\to vnv1\to ,v2\to ,\cdots ,vn\to .\; Allen\; will\; make nn moves.\; As\; Allen\text{'}s\; sense\; of\; direction\; is\; impaired,\; during\; the ii-th\; move\; he\; will\; either\; move\; in\; the\; direction \to vivi\to or -\to vi-vi\to .\; In\; other\; words,\; if\; his\; position\; is\; currently p=(x,y)p=(x,y),\; he\; will\; either\; move\; to p+\to vip+vi\to or p-\to vip-vi\to .}$

Allen doesn't want to wander too far from home (which happens to also be the bar). You need to help him figure out a sequence of moves (a sequence of signs for the vectors) such that his final position $\mathrm{pp satisfies |p|\le 1.5\cdot 106|p|\le 1.5\cdot 106 so\; that\; he\; can\; stay\; safe.}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; number\; of\; moves.}$

Each of the following lines contains two space-separated integers ${\mathrm{xixi and yiyi,\; meaning\; that \to vi=(xi,yi)vi\to =(xi,yi).\; We\; have\; that |vi|\le 106|vi|\le 106 for\; all ii.}}^{}$

Output

Output a single line containing $\mathrm{nn integers c1,c2,\cdots ,cnc1,c2,\cdots ,cn,\; each\; of\; which\; is\; either 11 or -1-1.\; Your\; solution\; is\; correct\; if\; the\; value\; of p=\sum ni=1ci\to vip=\sum i=1ncivi\to ,\; satisfies |p|\le 1.5\cdot 106|p|\le 1.5\cdot 106.}$

It can be shown that a solution always exists under the given constraints.

Examples

input

Copy

3
999999 0
0 999999
999999 0

output

Copy

1 1 -1 

input

Copy

1
-824590 246031

output

Copy

1 

input

Copy

8
-67761 603277
640586 -396671
46147 -122580
569609 -2112
400 914208
131792 309779
-850150 -486293
5272 721899

output

Copy

1 1 1 1 1 1 1 -1 

E题:
听他们说怎末瞎搞都能过>>>难受>>正解时说>>任意三个向量中的两个总可以合并成1个向量,并且长度不超过限制>>证明方法很巧妙
大佬们说这个题>>很难卡>>特别佩服杜教的random_shuffle>>>太强了>>收集了两个解法>>正解以后再补

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10; 
struct node{int x,y,v,id;}a[N];
typedef long long ll;
bool cmp(node a,node b){return abs(a.x-a.y)>abs(b.x-b.y);}
ll dis(ll x,ll y){return x*x+y*y;}
bool cmp1(node a,node b){return a.id<b.id;};
int main(){
    std::ios::sync_with_stdio(false);
    int n;cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i].x>>a[i].y,a[i].id=i;
    sort(a+1,a+n+1,cmp);
    ll X=0,Y=0;
    for(int i=1;i<=n;i++){
        if(dis(X+a[i].x,Y+a[i].y)<dis(X-a[i].x,Y-a[i].y)){
            X+=a[i].x,Y+=a[i].y;a[i].v=1;
        }
        else{
            X-=a[i].x,Y-=a[i].y;a[i].v=-1;
        }
    }
    sort(a+1,a+n+1,cmp1);
    for(int i=1;i<=n;i++) cout<<a[i].v<<" ";
    cout<<endl;
    return 0;
} 

View Code

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10; 
struct node{int x,y,id;}a[N];
typedef long long ll;
const ll tmp=1e6+5e5;
ll dis(ll x,ll y){return x*x+y*y;}
int ans[N];
int main(){
    std::ios::sync_with_stdio(false);
    int n;cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i].x>>a[i].y,a[i].id=i;
    while(1){
        random_shuffle(a+1,a+n+1);
        ll X=0,Y=0;
        for(int i=1;i<=n;i++){
            if(dis(X+a[i].x,Y+a[i].y)<dis(X-a[i].x,Y-a[i].y)){
                X+=a[i].x,Y+=a[i].y;ans[a[i].id]=1;
            }
            else{
                X-=a[i].x,Y-=a[i].y;ans[a[i].id]=-1;
            }
        }
        if(dis(X,Y)>=tmp*tmp) continue;
        for(int i=1;i<=n;i++) cout<<ans[i]<<" ";
        return 0;
    }
    return 0;
} 

View Code