本文解析了四道涉及字符串处理、二维数组分析、几何计算及动态规划的编程题。通过详细的代码实现，展示了如何解决实际问题，适用于算法初学者及竞赛选手。
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 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputLet'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 3 15 16 3 25 9
Output
3
Input
3 3 9 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 <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;
}
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