CodeForces-814C An impassioned circulation of affection

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.

Example
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 3 is 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
给你一串字符串，在输入k组数据，每组数据包含一个正整数p和一个字符ch，可以变换p个字符，使得字符串中
ch的连续序列最长，求最长连续序列。
题解

此题可以用尺取法来做，尺取法就是根据区间的特征交替推进左右端点求解问题。引用书上的话尺取法通常是对数组保存一对下标，即所选取的区间的左右端点，然后根据实际情况不断地推进区间左右端点以得出答案。之所以掌握这个技巧，是因为尺取法比直接暴力区间效率高的多，尤其是数据量大的时候，所以尺取法是一种高效的枚举区间的方法，一般用于求取有一定限制的区间个数或最短的区间。本题用尺取法一步步求ch的最长连续序列。

#include<stdio.h>                           //尺取法  构造区间逐渐开始寻找求最大值#include<string.h>#include<algorithm>using namespace std;int main(){    char a[1505],ch;    int n,i,j,lf,rg,k,maxx,p;    while(~scanf("%d",&n))    {        scanf("%s",a);        scanf("%d",&k);        while(k--)        {            scanf("%d %c",&p,&ch);            int num=0,len=0;            lf=0;                            //左端点            maxx=-1;            //printf("dggr %d\n",p);            for(rg=0; rg<n; rg++)            //右端点            {                if(a[rg]!=ch)                    num++;                   //不是ch的个数                len++;                // printf("mfoj   %d\n",num);                while(num>p)                //不是ch的字符个数大于输入的p                {                    if(a[lf]!=ch)                        num--;                                 lf++;                   //左端点向后移，下一步构造新的区间                    len--;                  //len--表示的是要求连续子序列都是ch的长度                }                // printf("******%d\n",len);                maxx=max(maxx,len);            }            printf("%d\n",maxx);        }    }    return 0;}