Codeforces 228 div2 A,B,C

全水题。。。。。

A. Fox and Number Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Fox Ciel is playing a game with numbers now.

Ciel has n positive integers: x1, x2, ..., xn. She can do the following operation as many times as needed: select two different indexes iand j such that xi > xj hold, and then apply assignment xi = xi - xj. The goal is to make the sum of all numbers as small as possible.

Please help Ciel to find this minimal sum.

Input

The first line contains an integer n (2 ≤ n ≤ 100). Then the second line contains n integers: x1, x2, ..., xn (1 ≤ xi ≤ 100).

Output

Output a single integer — the required minimal sum.

Sample test(s)

input

21 2

output

2

input

32 4 6

output

6

input

212 18

output

12

input

545 12 27 30 18

output

15

Note

In the first example the optimal way is to do the assignment: x2 = x2 - x1.

In the second example the optimal sequence of operations is: x3 = x3 - x2, x2 = x2 - x1.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int num[1200],n;int gcd(int a,int b){return b==0?a:gcd(b,a%b);}int main(){scanf("%d",&n);int a=-1;for(int i=0;i<n;i++) {scanf("%d",&num[i]);if(a==-1) a=num[i];else a=gcd(a,num[i]);}printf("%d\n",a*n);return 0;}

B. Fox and Cross

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Fox Ciel has a board with n rows and n columns. So, the board consists of n × n cells. Each cell contains either a symbol '.', or a symbol '#'.

A cross on the board is a connected set of exactly five cells of the board that looks like a cross. The picture below shows how it looks.

Ciel wants to draw several (may be zero) crosses on the board. Each cross must cover exactly five cells with symbols '#', and any cell with symbol '#' must belong to some cross. No two crosses can share a cell.

Please, tell Ciel if she can draw the crosses in the described way.

Input

The first line contains an integer n (3 ≤ n ≤ 100) — the size of the board.

Each of the next n lines describes one row of the board. The i-th line describes the i-th row of the board and consists of n characters. Each character is either a symbol '.', or a symbol '#'.

Output

Output a single line with "YES" if Ciel can draw the crosses in the described way. Otherwise output a single line with "NO".

Sample test(s)

input

5.#...####..####...#......

output

YES

input

4################

output

NO

input

6.#....####...####..#.##.######.#..#.

output

YES

input

6.#..#.######.####..####.######.#..#.

output

NO

input

3.........

output

YES

Note

In example 1, you can draw two crosses. The picture below shows what they look like.

In example 2, the board contains 16 cells with '#', but each cross contains 5. Since 16 is not a multiple of 5, so it's impossible to cover all.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n,nb;char g[1000][1000];bool ck(int x,int y){if((x>=0&&x<n)&&(y>=0&&y<n)) return true;return false;}void dfs(int x,int y){if(ck(x+1,y)&&ck(x+2,y)&&ck(x+1,y-1)&&ck(x+1,y+1)){if(g[x][y]=='#'&&g[x+1][y]=='#'&&g[x+1][y-1]=='#'&&g[x+1][y+1]=='#'&&g[x+2][y]=='#'){g[x][y]=g[x+1][y]=g[x+1][y-1]=g[x+1][y+1]=g[x+2][y]='X';nb+=5;}}}int main(){scanf("%d",&n);for(int i=0;i<n;i++){scanf("%s",g[i]);}int jinghao=0;for(int i=0;i<n;i++){for(int j=0;j<n;j++){if(g[i][j]=='#') jinghao++;}}if(jinghao%5) {puts("NO"); return 0;}for(int i=0;i<n;i++){for(int j=0;j<n;j++){if(g[i][j]=='#'){dfs(i,j);}}}if(nb!=jinghao){puts("NO");}else{puts("YES");}return 0;}

A. Fox and Box Accumulation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).

Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.

Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more thanxi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?

Input

The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).

Output

Output a single integer — the minimal possible number of piles.

Sample test(s)

input

30 0 10

output

2

input

50 1 2 3 4

output

1

input

40 0 0 0

output

4

input

90 1 0 2 0 1 1 2 10

output

3

Note

In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.

In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int st[1000],n,ans=0;int vis[1000];int main(){scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d",st+i);sort(st,st+n);for(int i=0;i<n;i++){if(vis[i]==0){ans++;int strength=1;vis[i]=ans;for(int j=i+1;j<n;j++){if(vis[j]) continue;if(st[j]>=strength){vis[j]=ans;strength++;}} }}printf("%d\n",ans);return 0;}