您的当前位置：首页 Codeforces#418C An impassioned circulation of affection

Codeforces#418C An impassioned circulation of affection

来源：客趣旅游网

Nadeko's birthday is approaching! As she decorated the room for the party, a long garland of Dianthus-shaped paper pieces was placed on a prominent part of the wall. Brother Koyomi will like it!

Still unsatisfied with the garland, Nadeko decided to polish it again. The garland has n pieces numbered from 1 to n from left to right, and the i-th piece has a colour si, denoted by a lowercase English letter. Nadeko will repaint at most m of the pieces to give each of them an arbitrary new colour (still denoted by a lowercase English letter). After this work, she finds out all subsegments of the garland containing pieces of only colour c — Brother Koyomi's favourite one, and takes the length of the longest among them to be the Koyomity of the garland.

For instance, let's say the garland is represented by "kooomo", and Brother Koyomi's favourite colour is "o". Among all subsegments containing pieces of "o" only, "ooo" is the longest, with a length of 3. Thus the Koyomity of this garland equals 3.

But problem arises as Nadeko is unsure about Brother Koyomi's favourite colour, and has swaying ideas on the amount of work to do. She has q plans on this, each of which can be expressed as a pair of an integer mi and a lowercase letter ci, meanings of which are explained above. You are to find out the maximum Koyomity achievable after repainting the garland according to each plan.

Input

The first line of input contains a positive integer n (1 ≤ n ≤ 1 500) — the length of the garland.

The second line contains n lowercase English letters s1s2... sn as a string — the initial colours of paper pieces on the garland.

The third line contains a positive integer q (1 ≤ q ≤ 200 000) — the number of plans Nadeko has.

The next q lines describe one plan each: the i-th among them contains an integer mi (1 ≤ mi ≤ n) — the maximum amount of pieces to repaint, followed by a space, then by a lowercase English letter ci — Koyomi's possible favourite colour.

Output

Output q lines: for each work plan, output one line containing an integer — the largest Koyomity achievable after repainting the garland according to it.

   Examples 
 

     input 
   
6
koyomi
3
1 o
4 o
4 m

     output 
   
3
6
5

     input 
   
15
yamatonadeshiko
10
1 a
2 a
3 a
4 a
5 a
1 b
2 b
3 b
4 b
5 b

     output 
   
3
4
5
7
8
1
2
3
4
5

     input 
   
10
aaaaaaaaaa
2
10 b
10 z

     output 
   
10
10

Note

In the first sample, there are three plans:

In the first plan, at most 1 piece can be repainted. Repainting the "y" piece to become "o" results in "kooomi", whose Koyomity of 3is the best achievable;
In the second plan, at most 4 pieces can be repainted, and "oooooo" results in a Koyomity of 6;
In the third plan, at most 4 pieces can be repainted, and "mmmmmi" and "kmmmmm" both result in a Koyomity of 5.

题意：给你一个字符串，问你，最多改k个，且只能改为一种字符，最长连续相同该种字符的长度是多少。

暴力预处理下就好

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
int dp[1505][30];
char str[1505];
int main()
{
    int n, q, num;
    while(cin >> n)
    {
        scanf("%s", str);
        memset(dp, -1, sizeof(dp));
        for(int ch = 0; ch < 26; ch++)
        {
            for(int i = 0; i < n; i++)
            {
                num = 0;
                for(int j = i; j >= 0; j--)
                {
                    if(str[j] == (char)('a'+ch)) num++;
                    dp[i-j+1-num][ch] = max(dp[i-j+1-num][ch], i - j+1);
                }
            }
        }
        char ch;
        scanf("%d", &q);
        for(int i = 1; i <= q; i++)
        {
            scanf("%d %c", &num, &ch);
            if(dp[num][ch-'a'] == -1)
                printf("%d\n", n);
            else
                printf("%d\n", dp[num][ch-'a']);
        }
    }
    return 0;
}

因篇幅问题不能全部显示，请点此查看更多更全内容

查看全文