Educational Codeforces Round 26 A B C 三道水题

A：

#include <iostream>#include <cstdio>#include <string.h>#include <algorithm>using namespace std;int main(){int ans = 0;int n;cin>>n;char c;int maxn = 0;getchar();while( n-- ){c = getchar();if( c >= 'A' && c <= 'Z' ){ans++;}if( c == ' ' || n == 0 ) { maxn = max(maxn,ans); ans = 0;}//cout<<"ans ; "<<ans<<endl;}printf("%d\n",maxn);return 0;}

B：

#include <iostream>#include <cstdio>#include <string.h>#include <map>#include <algorithm>using namespace std;const int AX = 200+2;char mp[AX][AX];map<char,int>p;int main(){int n,m;cin>>n>>m;for( int i = 0 ; i < n ; i++ ){scanf("%s",mp[i]);}int flagm = 1;int flagn = 1;if( m % 3 && n % 3 ) {printf("NO\n");return 0 ;}if( n == 3  && m == 1){if( mp[0][0] != mp[1][0] && mp[0][0] != mp[2][0] && mp[1][0] != mp[2][0] ){printf("YES\n");return 0;}  else {printf("NO\n");return 0;}}else if( m == 3  && n == 1){if( mp[0][0] != mp[0][1] && mp[0][0] != mp[0][2] && mp[0][1] != mp[0][2] ){printf("YES\n");return 0 ;} else {printf("NO\n");return 0;}}if( m == 1 && n > 3 && n % 3 == 0 ){int ave = n/3;for( int i = 0; i < n ;i += ave ){for(int j = i+1 ; j < i+ave; j++ ){if( mp[j][0] != mp[j-1][0] ) {flagn = 0;break;}p[mp[j][0]] = 1;}if(flagn == 0 ) break;} if( !p['R'] || !p['G'] || !p['B'] ) flagn = 0;}else if( n == 1 && m>3 && m % 3 == 0 ){int ave = m/3;for( int i = 0; i < m ;i += ave ){for(int j = i+1 ; j < i+ave; j++ ){if( mp[0][j] != mp[0][j-1] ) {flagm = 0;break;}p[mp[0][j]] = 1;}if(flagm == 0 ) break;} if( !p['R'] || !p['G'] || !p['B'] ) flagm = 0;}else{if( m % 3 == 0 ){int ave = m/3;for( int j = 0,num = 1 ;j < m ; j++,num++ ){if( num  == ave + 1 ) { p[mp[1][j-1]] = 1;num = 1; }for( int i = 1 ; i < n  ; i ++ ){if( mp[i][j] != mp[i-1][j] || p[mp[i][j]] )  {flagm = 0 ; break;}}if( flagm == 0  ) break;}}if( n % 3 == 0 ){int ave = n/3;for( int i = 0,num = 1 ;i < n ; i++,num++ ){if( num  == ave + 1 ) { p[mp[i-1][1]] = 1; num = 1;}for( int j = 1 ; j < m  ; j++ ){if( mp[i][j] != mp[i][j-1] || p[mp[i][j]] )  { flagn = 0 ; break;}}if( flagn == 0  ) break;}}}if( m % 3 ==0 && n % 3 == 0  ){if( flagn || flagm ) printf("YES\n");else printf("NO\n");}else if( m % 3 == 0 ){if( flagm ) printf("YES\n");else printf("NO\n");}else{if( flagn ) printf("YES\n");else printf("NO\n");}return 0;}

C：

#include <iostream>#include <cstdio>using namespace std;const int AX = 1e2+3;struct Node{int x;int y;}s[AX];int main(){int n,a,b;int cnt = 0;scanf("%d%d%d",&n,&a,&b);int x1,y1;for( int i = 0 ; i < n ; i++ ){scanf("%d%d",&x1,&y1);if( ( y1 > a && y1 > b ) || ( x1 > a && x1 > b ) || (x1 == a && y1 == b) || (y1 == a && x1 == b ) ) continue;s[cnt].x = x1;s[cnt].y  = y1;cnt++;}/*for( int i = 0 ; i < cnt ; i ++ ){cout<<"s[]: "<<s[i].x<<' '<<" s[] "<<s[i].y<<endl;}*/int ans = 0;for( int i = 0 ; i < cnt ; i++ ){for( int j = 0 ; j < cnt ; j++ ){if( i == j ) continue;if( s[i].x + s[j].y <= a && s[i].y <=b && s[j].x <=b ) { ans = max( ans , s[i].x * s[i].y + s[j].x * s[j].y ); continue;}if( s[i].x + s[j].y <= b && s[i].y <=a && s[j].x <=a ) { ans = max( ans , s[i].x * s[i].y + s[j].x * s[j].y ); continue;}if( s[i].x + s[j].x <= a && s[i].y <=b && s[j].y <=b ) { ans = max( ans , s[i].x * s[i].y + s[j].x * s[j].y ); continue;}if( s[i].x + s[j].x <= b && s[i].y <=a && s[j].y <=a ) { ans = max( ans , s[i].x * s[i].y + s[j].x * s[j].y ); continue;}if( s[i].y + s[j].x <= a && s[i].x <=b && s[j].y <=b ) { ans = max( ans , s[i].x * s[i].y + s[j].x * s[j].y ); continue;}if( s[i].y + s[j].x <= b && s[i].x <=a && s[j].y <=a ) { ans = max( ans , s[i].x * s[i].y + s[j].x * s[j].y ); continue;}if( s[i].y + s[j].y <= b && s[i].x <=a && s[j].x <=a ) { ans = max( ans , s[i].x * s[i].y + s[j].x * s[j].y ); continue;}if( s[i].y + s[j].y <= a && s[i].x <=b && s[j].x <=b ) { ans = max( ans , s[i].x * s[i].y + s[j].x * s[j].y ); continue;}}}cout<<ans<<endl;return 0;}

D：

D. Round Subset

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's call the roundness of the number the number of zeros to which it ends.

You have an array of n numbers. You need to choose a subset of exactlyk numbers so that the roundness of the product of the selected numbers will be maximum possible.

Input

The first line contains two integer numbers n andk (1 ≤ n ≤ 200, 1 ≤ k ≤ n).

The second line contains n space-separated integer numbersa₁, a₂, ..., a_n (1 ≤ a_i ≤ 10¹⁸).

Output

Print maximal roundness of product of the chosen subset of length k.

Examples

Input

3 250 4 20

Output

3

Input

5 315 16 3 25 9

Output

3

Input

3 39 77 13

Output

0

Note

In the first example there are 3 subsets of2 numbers. [50, 4] has product200 with roundness2, [4, 20] — product80, roundness1, [50, 20] — product1000, roundness3.

In the second example subset [15, 16, 25] has product6000, roundness3.

In the third example all subsets has product with roundness 0.

dp[i][j]表示选i个数中 5的个数，j代表 2的个数

#include <bits/stdc++.h>#define LL long long#define INF 0x7777777using namespace std;const int AX = 200+6;const int maxn = 64*AX;LL a[AX];int dp[AX][maxn];int main(){int n,k;while( ~scanf("%d%d",&n,&k) ){memset( a , 0 , sizeof(a) );LL x;for( int i = 0 ; i < n ; i++ ){cin>>a[i];}for( int i = 0 ; i <= k ; i++ ){for( int j = 0 ; j < maxn ; j++ ){dp[i][j] = -INF;}}dp[0][0] = 0;for( int i = 0 ; i < n ; i++ ){LL temp = a[i];int num2 = 0 , num5 = 0 ;while( temp % 5  == 0 ){num5 ++;temp /= 5;}temp = a[i];while( temp % 2 == 0 ){num2 ++;temp /= 2;}for (int j = k ; j >= 1; j-- ){for( int l = num2 ; l < maxn ;l++ ){dp[j][l] = max( dp[j-1][l-num2] + num5 ,dp[j][l] );}}}int res = 0;for( int i = 1 ; i < maxn ; i++ ){res = max( res , min( i , dp[k][i] ) );}printf("%d\n",res);}return 0;}