codeforces 877b
2 seconds
256 megabytes
standard input
standard output
One day Nikita found the string containing letters "a" and "b" only.
Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters "a" and the 2-nd contains only letters "b".
Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get?
The first line contains a non-empty string of length not greater than 5 000 containing only lowercase English letters "a" and "b".
Print a single integer — the maximum possible size of beautiful string Nikita can get.
abba
4
bab
2
It the first sample the string is already beautiful.
In the second sample he needs to delete one of "b" to make it beautiful.
思路：记录每个位置后面有多少个a，有多少个b 存在arr[5005][2]里面 arr[i][0]表示i后面包括i有arr[i][0]个a arr[i][1]表示i后面包括i有arr[i][1]个b
遍历每个段位
for(i =0 ; i < strlen(str);++i)
{
if(str[j] == 'b')
{
for(j = i; j < strlen(str);++j)
{
if(str[j] == 'b')
maxn = max(maxn, arr[0][0] - arr[i][0] + arr[i][2] - arr[j][2] + 1 + arr[j][0]);
}
}
}
注意全是a的情况
丑陋的代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
int main()
{
// char str[5010];
string str;
int arr[5010][2];
int i,j,k,a,b,a1,b1;
// memset(str,0,sizeof(str));
int maxn = 0;
a1 = b1 = a = b = 0;
// scanf("%s",str);
cin >> str;
memset(arr, 0, sizeof(arr));
for(i = 0; i < str.length(); ++i)
{
if(str[i] == 'a')
a++;
else
b++;
}
if(b == 0)
{
printf("%d\n",(int)str.length());
return 0;
}
a1 = b1 = 0;
for(i = 0; i < str.length(); ++i)
{
if(str[i] == 'a')
{
arr[i][0] = a - a1;
arr[i][1] = b - b1;
a1++;
}
else
{
arr[i][0] = a - a1;
arr[i][1] = b - b1;
b1++;
}
}
a1 = b1 = 0;
for(i = 0; i < str.length(); ++i)
{
if(str[i] == 'b')
{
for(j = i; j < str.length(); ++j)
{
if(str[j] == 'b')
maxn = max(maxn,a - arr[i][0] + arr[i][1] - arr[j][1] + arr[j][0]+1);
}
}
}
printf("%d\n",maxn);
}
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