题目链接
https://acm.bnu.edu.cn/v3/contest_show.php?cid=8506#problem/A
problem description
As we know, the NTU Final PK contest usually tends to be pretty hard. Many teams got frustrated when participating NTU Final PK contest. So I decide to make the first problem as “easy” as possible. But how to know how easy is a problem? To make our life easier, we just consider how easy is a string. Here, we introduce a sane definition of “easiness”. The easiness of a string is the maximum times of “easy” as a subsequence of it. For example, the easiness of “eeaseyaesasyy” is 2. Since “easyeasy” is a subsequence of it, but “easyeasyeasy” is too easy. How to calculate easiness seems to be very easy. So here is a string s consists of only ‘e’, ‘a’, ‘s’, and ‘y’. Please answer m queries. The i-th query is a interval [li , ri ], and please calculate the easiness of s[li ..ri ].
Input
The first line contains a string s. The second line contains an integer m. Each of following m lines contains two integers li , ri . • 1 ≤ |s| ≤ 105 • 1 ≤ m ≤ 105 • 1 ≤ li ≤ ri ≤ |s| • s consists of only ‘e’, ‘a’, ‘s’, and ‘y’
Output
For each query, please output the easiness of that substring in one line.
Examples
standard input
easy
3
1 4
2 4
1 3
eeaseyaesasyy
4
1 13
2 12
2 10
3 11
standard output
1
0
0
2
2
1
0
题意：给了一个只含有'e' 'a' 's' 'y' 的字符串然后m次询问，每次询问输入l r 求这个区间含有多少个“easy”序列（每个“easy” 字符之间不需要连在一起）；
思路：用倍增的思路来做，每个点只记录最靠近它的在它左边的那个字母的位置，比如'e'记录前面的'y','a'记录前面的'e','s'记录前面的'a'，'y'记录前面的's' 并注意记录距离i最近（左边的y）的y的位置（用p[i]存储） 定义anc[i][j] 表示第i个字符前的第（1<<j）个字符的位置，这个可以用倍增做到 anc[i][j]=anc[anc[i][j-1]][j-1]， 查询时，先找到v=p[r] 然后找左边有效字符个数，最后除以4 就是结果；
代码如下：
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <cstring>
#include <queue>
using namespace std;
typedef long long LL;
char s[];
int a[],p[];
int anc[][];
int mp[]; int main()
{
int m;
while(scanf("%s",s+)!=EOF)
{
int len=strlen(s+);
for(int i=;i<=len;i++)
{
if(s[i]=='e') a[i]=;
else if(s[i]=='a') a[i]=;
else if(s[i]=='s') a[i]=;
else a[i]=;
}
memset(mp,,sizeof(mp));
for(int i=;i<=len;i++)
{
int pre=(a[i]-+)%;
anc[i][]=mp[pre];
mp[a[i]]=i;
p[i]=mp[];
}
for(int i=;i<=;i++)
{
for(int j=;j<=len;j++)
{
anc[j][i]=anc[anc[j][i-]][i-];
}
}
scanf("%d",&m);
while(m--)
{
int l,r;
int sum=;
scanf("%d%d",&l,&r);
int v=p[r];
for(int i=;i>=;i--)
{
if(anc[v][i]>=l){
sum+=(<<i);
v=anc[v][i];
}
}
printf("%d\n",sum/);
}
}
return ;
}