Codeforces Round #483 (Div. 2)题解

A. Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Two players play a game.

Initially there are $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an written\; on\; the\; board.\; Each\; turn\; a\; player\; selects\; one\; number\; and\; erases\; it\; from\; the\; board.\; This\; continues\; until\; there\; is\; only\; one\; number\; left\; on\; the\; board,\; i. e. n-1n-1 turns\; are\; made.\; The\; first\; player\; makes\; the\; first\; move,\; then\; players\; alternate\; turns.}$

The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.

You want to know what number will be left on the board after $\mathrm{n-1n-1 turns\; if\; both\; players\; make\; optimal\; moves.}$

Input

The first line contains one integer $\mathrm{nn (1\le n\le 10001\le n\le 1000) —\; the\; number\; of\; numbers\; on\; the\; board.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1061\le ai\le 106).}$

Output

Print one number that will be left on the board.

Examples

input

Copy

3
2 1 3

output

Copy

2

input

Copy

3
2 2 2

output

Copy

2

Note

In the first sample, the first player erases $\mathrm{33 and\; the\; second\; erases 11. 22 is\; left\; on\; the\; board.}$

In the second sample, $\mathrm{22 is\; left\; on\; the\; board\; regardless\; of\; the\; actions\; of\; the\; players.}$

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                                                                        .                                                                             **/
                                                                        
#include <bits/stdc++.h>
using namespace std;

#define FFI(a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 1006
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; 

int main(void){
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(false); cin.tie(0);
    int n, num[MAX] = {0};
    cin>>n;
    FFI(1, n + 1) cin>>num[i];
    sort(num, num + n + 1);
    if(n & 1) cout<<num[n / 2 + 1]<<endl;
    else cout<<num[n / 2]<<endl;        
    return EXIT_SUCCESS;
}

 

B. Minesweeper

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

 

One day Alex decided to remember childhood when computers were not too powerful and lots of people played only default games. Alex enjoyed playing Minesweeper that time. He imagined that he saved world from bombs planted by terrorists, but he rarely won.

Alex has grown up since then, so he easily wins the most difficult levels. This quickly bored him, and he thought: what if the computer gave him invalid fields in the childhood and Alex could not win because of it?

He needs your help to check it.

A Minesweeper field is a rectangle $\mathrm{n\times mn\times m,\; where\; each\; cell\; is\; either\; empty,\; or\; contains\; a\; digit\; from 11 to 88,\; or\; a\; bomb.\; The\; field\; is\; valid\; if\; for\; each\; cell:}$

• if there is a digit $\mathrm{kk in\; the\; cell,\; then\; exactly kk neighboring\; cells\; have\; bombs.}$
• if the cell is empty, then all neighboring cells have no bombs.

Two cells are neighbors if they have a common side or a corner (i. e. a cell has at most $\mathrm{88 neighboring\; cells).}$

 

Input

The first line contains two integers $\mathrm{nn and mm (1\le n,m\le 1001\le n,m\le 100)\; —\; the\; sizes\; of\; the\; field.}$

The next $\mathrm{nn lines\; contain\; the\; description\; of\; the\; field.\; Each\; line\; contains mm characters,\; each\; of\; them\; is\; "."\; (if\; this\; cell\; is\; empty),\; "*"\; (if\; there\; is\; bomb\; in\; this\; cell),\; or\; a\; digit\; from 11 to 88,\; inclusive.}$

 

Output

Print "YES", if the field is valid and "NO" otherwise.

You can choose the case (lower or upper) for each letter arbitrarily.

 

Examples

input

Copy

3 3
111
1*1
111

output

Copy

YES

input

Copy

2 4
*.*.
1211

output

Copy

NO

 

Note

In the second example the answer is "NO" because, if the positions of the bombs are preserved, the first line of the field should be *2*1.

You can read more about Minesweeper in Wikipedia's article.

分析：这题数太小，答案出错就是数字不对和*周围有.，所以暴力分析一遍即可

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define MAX 110
#define ME(a, b) memset(a, b, sizeof(a))

int dir[8][2] = {-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};    //8个方向 
char matrix[MAX][MAX];
int main(void){
    ios::sync_with_stdio(false);cin.tie(0);
    int i, j, k, n, m, temp;
    
    cin>>n>>m;
    for(i = 1; i <= n; i++)for(j = 1; j <= m; j++)cin>>matrix[i][j];
    for(i = 1; i <= n; i++){
        for(j = 1; j <= m; j++){
            if(matrix[i][j] == '*'){
                for(k = 0; k < 8; k++){
                    int a = i + dir[k][0];
                    int b = j + dir[k][1];
                    if(matrix[a][b] == '.'){
                        cout<<"NO"<<endl;
                        return 0;
                    }
                }
            }
            else if(matrix[i][j] >= '0' && matrix[i][j] <= '9'){
                temp = matrix[i][j] - '0';
                for(k = 0; k < 8; k++){
                    int a = i + dir[k][0];
                    int b = j + dir[k][1];
                    if(matrix[a][b] == '*') temp--;
                }
                if(temp != 0){
                    cout<<"NO"<<endl;
                    return 0;
                }
            }
        }
    }
    cout<<"YES"<<endl;
    
    return EXIT_SUCCESS;
}

 

C. Finite or not?

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given several queries. Each query consists of three integers $\mathrm{pp, qq and bb.\; You\; need\; to\; answer\; whether\; the\; result\; of p/qp/q in\; notation\; with\; base bb is\; a\; finite\; fraction.}$

A fraction in notation with base $\mathrm{bb is\; finite\; if\; it\; contains\; finite\; number\; of\; numerals\; after\; the\; decimal\; point.\; It\; is\; also\; possible\; that\; a\; fraction\; has\; zero\; numerals\; after\; the\; decimal\; point.}$

Input

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

Next $\mathrm{nn lines\; contain\; queries,\; one\; per\; line.\; Each\; line\; contains\; three\; integers pp, qq,\; and bb (0\le p\le 10180\le p\le 1018, 1\le q\le 10181\le q\le 1018, 2\le b\le 10182\le b\le 1018).\; All\; numbers\; are\; given\; in\; notation\; with\; base 1010.}$

Output

For each question, in a separate line, print Finite if the fraction is finite and Infinite otherwise.

Examples

input

Copy

2
6 12 10
4 3 10

output

Copy

Finite
Infinite

input

Copy

4
1 1 2
9 36 2
4 12 3
3 5 4

output

Copy

Finite
Finite
Finite
Infinite

Note

$\frac{\mathrm{612=12=0,510612=12=0,510}}{}$

$\frac{\mathrm{43=1,(3)1043=1,(3)10}}{}$

$\frac{\mathrm{936=14=0,012936=14=0,012}}{}$

$\frac{\mathrm{412=13=0,13}}{}$

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

int main(void){
    LL n, a, b, c;
    ios::sync_with_stdio(false);cin.tie(0);
    
    cin>>n;
    while(n--){
        cin>>a>>b>>c;
        b /= __gcd(a, b);    //b转换成最简分母 
        c = __gcd(b, c);
        while(c != 1){    //可化成c进制的小数 
            b /= __gcd(b, c);    //b转换成c进制 
            c = __gcd(b, c);
        }
        if(b == 1) cout<<"Finite"<<endl;
        else cout<<"Infinite"<<endl;
    }
    return EXIT_SUCCESS;
}

 

D. XOR-pyramid

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

For an array $\mathrm{bb of\; length mm we\; define\; the\; function ff as}$

$\mathrm{f(b)=\{b[1]if m=1f(b[1]\oplus b[2],b[2]\oplus b[3],\dots ,b[m-1]\oplus b[m])otherwise,f(b)=\{b[1]if m=1f(b[1]\oplus b[2],b[2]\oplus b[3],\dots ,b[m-1]\oplus b[m])otherwise,}$

where $\oplus \oplus is bitwise\; exclusive\; OR.$

For example, $\mathrm{f(1,2,4,8)=f(1\oplus 2,2\oplus 4,4\oplus 8)=f(3,6,12)=f(3\oplus 6,6\oplus 12)=f(5,10)=f(5\oplus 10)=f(15)=15f(1,2,4,8)=f(1\oplus 2,2\oplus 4,4\oplus 8)=f(3,6,12)=f(3\oplus 6,6\oplus 12)=f(5,10)=f(5\oplus 10)=f(15)=15}$

You are given an array $\mathrm{aa and\; a\; few\; queries.\; Each\; query\; is\; represented\; as\; two\; integers ll and rr.\; The\; answer\; is\; the\; maximum\; value\; of ff on\; all\; continuous\; subsegments\; of\; the\; array al,al+1,\dots ,aral,al+1,\dots ,ar.}$

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 50001\le n\le 5000) —\; the\; length\; of aa.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (0\le ai\le 230-10\le ai\le 230-1) —\; the\; elements\; of\; the\; array.}$

The third line contains a single integer $\mathrm{qq (1\le q\le 1000001\le q\le 100000) —\; the\; number\; of\; queries.}$

Each of the next $\mathrm{qq lines\; contains\; a\; query\; represented\; as\; two\; integers ll, rr (1\le l\le r\le n1\le l\le r\le n).}$

Output

Print $\mathrm{qq lines —\; the\; answers\; for\; the\; queries.}$

Examples

input

Copy

3
8 4 1
2
2 3
1 2

output

Copy

5
12

input

Copy

6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2

output

Copy

60
30
12
3

Note

In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.

In second sample, optimal segment for first query are $[3,6][3,6],\; for\; second\; query\; — [2,5][2,5],\; for\; third\; — [3,4][3,4],\; for\; fourth\; — [1,2][1,2].$

 

/**
 
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                                                                        .                                                                             **/
                                                                        
#include <bits/stdc++.h>
using namespace std;

#define FFI(a, b) for(int i = a; i < b; i++)
#define FFJ(a, b) for(int j = a; j < b; j++)
#define RR(a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 5005
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; 

LL num[MAX];
LL dp[MAX][MAX];

inline int max(int a, int b){return a > b ? a : b;}
int main(void){
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(false); cin.tie(0);
    int i, j, n, q, l, r, ma;
    
    cin>>n;
    FFI(0, n){
        cin>>num[i];
        dp[i][i] = num[i];
    }
    FFI(1, n) FFJ(0, n - i) dp[j][j + i] = dp[j][j + i - 1] xor dp[j + 1][j + i];    //斜三角
    FFI(1, n) FFJ(0, n - i) dp[j][j + i] = max(dp[j][j + i], max(dp[j][j + i - 1], dp[j + 1][j + i]));

    cin>>q;
    while(q--){
        cin>>l>>r;
        l--;r--;
        cout<<dp[l][r]<<endl;
    }     
    
    return EXIT_SUCCESS;
}

 

E. Elevator

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You work in a big office. It is a 9 floor building with an elevator that can accommodate up to 4 people. It is your responsibility to manage this elevator.

Today you are late, so there are queues on some floors already. For each person you know the floor where he currently is and the floor he wants to reach. Also, you know the order in which people came to the elevator.

According to the company's rules, if an employee comes to the elevator earlier than another one, he has to enter the elevator earlier too (even if these employees stay on different floors). Note that the employees are allowed to leave the elevator in arbitrary order.

The elevator has two commands:

• Go up or down one floor. The movement takes 1 second.
• Open the doors on the current floor. During this operation all the employees who have reached their destination get out of the elevator. Then all the employees on the floor get in the elevator in the order they are queued up while it doesn't contradict the company's rules and there is enough space in the elevator. Each employee spends 1 second to get inside and outside the elevator.

Initially the elevator is empty and is located on the floor 1.

You are interested what is the minimum possible time you need to spend to deliver all the employees to their destination. It is not necessary to return the elevator to the floor 1.

Input

The first line contains an integer n (1 ≤ n ≤ 2000) — the number of employees.

The i-th of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 9, ai ≠ bi) — the floor on which an employee initially is, and the floor he wants to reach.

The employees are given in the order they came to the elevator.

Output

Print a single integer — the minimal possible time in seconds.

Examples

input

Copy

2
3 5
5 3

output

Copy

10

input

Copy

2
5 3
3 5

output

Copy

12
分析：这题最JB有意思，DP[a][b][c][d][e][f]装小时间，a,b,c,d是几层，f是向上走，还是想下走，很实用啊

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                                                                        .                                                                             **/
                                                                        
#include <bits/stdc++.h>
using namespace std;

#define FFI(a, b) for(int i = a; i < b; i++)
#define FFJ(a, b) for(int j = a; j < b; j++)
#define RR(a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 1001
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; 


int n;
int a[2001],b[2001];
int dp[2002][10][10][10][10][2];    //n个点到4个方向的上下值 
int Solve(int ind,int x1, int x2, int x3, int x4, int y){
    if(dp[ind][x1][x2][x3][x4][y] != -1) return dp[ind][x1][x2][x3][x4][y];
    int ans=32000, c1 = ind, c2 = x4, cur;
    if(y==1){
        cur=a[ind];
        x4=b[ind];
        ind++;
    }
    else{
        cur=x4;
        x4=0;
    }
    if(ind>n && x1==0 && x2==0 && x3==0 && x4==0)
        return 0;
    if(x1!=0)
        ans=min(ans,Solve(ind,x2,x3,x4,x1,0)+max(cur-x1,x1-cur)+1);
    if(x2!=0)
        ans=min(ans,Solve(ind,x1,x3,x4,x2,0)+max(cur-x2,x2-cur)+1);
    if(x3!=0)
        ans=min(ans,Solve(ind,x1,x2,x4,x3,0)+max(cur-x3,x3-cur)+1);
    if(x4!=0)
        ans=min(ans,Solve(ind,x1,x2,x3,x4,0)+max(cur-x4,x4-cur)+1);
    if(x1==0 && ind<=n)
        ans=min(ans,Solve(ind,x2,x3,x4,x1,1)+max(cur-a[ind],a[ind]-cur)+1);
    else if(x2==0 && ind<=n)
        ans=min(ans,Solve(ind,x1,x3,x4,x2,1)+max(cur-a[ind],a[ind]-cur)+1);
    else if(x3==0 && ind<=n)
        ans=min(ans,Solve(ind,x1,x2,x4,x3,1)+max(cur-a[ind],a[ind]-cur)+1);
    else if(x4==0 && ind<=n)
        ans=min(ans,Solve(ind,x1,x2,x3,x4,1)+max(cur-a[ind],a[ind]-cur)+1);
    return dp[c1][x1][x2][x3][c2][y]=ans;
}
int main(void){
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(false); cin.tie(0);
    
    cin>> n;
    for(int i = 1;i <= n + 1; i++){
        if(i <= n) cin>>a[i]>>b[i];
        for(int j = 0; j <= 9; j++)
            for(int k = 0; k <= 9; k++)
                for(int x = 0; x <= 9; x++)
                    for(int y = 0; y <= 9; y++)
                        for(int z = 0; z <= 1; z++)
                            dp[i][j][k][x][y][z] = -1;
    }
    cout<<Solve(1, 0, 0, 0, 0, 1) + a[1]<<endl;
    return EXIT_SUCCESS;
}