Educational Codeforces Round 26 A B C 三道水题 - - D 动态规划

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 exactly
k 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 and
k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).

The second line contains n space-separated integer numbers
a1, a2, ..., an
(1 ≤ ai ≤ 1018).

Output

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

Examples

Input

3 2
50 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 of
2 numbers. [50, 4] has product
200 with roundness
2, [4, 20] — product
80, roundness
1, [50, 20] — product
1000, roundness
3.

In the second example subset [15, 16, 25] has product
6000, roundness
3.

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 <
4000;bits/stdc++.h>
#define LL long long
#define INF 0x7777777
using 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;
}