2016弱校聯盟十一專場10.5---As Easy As Possible（倍增）

題目鏈接

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

題意：給了一個只含有'e'  'a'  's'  'y'  的字元串然後m次詢問，每次詢問輸入l r 求這個區間含有多少個“easy”序列（每個“easy” 字元之間不需要連在一起）；

思路：用倍增的思路來做，每個點只記錄最靠近它的在它左邊的那個字母的位置，比如'e'記錄前面的'y','a'記錄前面的'e','s'記錄前面的'a'，'y'記錄前面的's' 並註意記錄距離i最近（左邊的y）的y的位置（用p[i]存儲）  定義anc[i][j] 表示第i個字元前的第（1<<j）個字元的位置，這個可以用倍增做到 anc[i][j]=anc[anc[i][j-1]][j-1]，  查詢時，先找到v=p[r] 然後找左邊有效字元個數，最後除以4 就是結果；

代碼如下：

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <cstring>
#include <queue>
using namespace std;
typedef long long LL;
char s[100005];
int a[100005],p[100005];
int anc[100005][21];
int mp[4];

int main()
{
    int m;
    while(scanf("%s",s+1)!=EOF)
    {
        int len=strlen(s+1);
        for(int i=1;i<=len;i++)
        {
            if(s[i]=='e') a[i]=0;
            else if(s[i]=='a') a[i]=1;
            else if(s[i]=='s') a[i]=2;
            else a[i]=3;
        }
        memset(mp,0,sizeof(mp));
        for(int i=1;i<=len;i++)
        {
            int pre=(a[i]-1+4)%4;
            anc[i][0]=mp[pre];
            mp[a[i]]=i;
            p[i]=mp[3];
        }
        for(int i=1;i<=20;i++)
        {
            for(int j=1;j<=len;j++)
            {
                anc[j][i]=anc[anc[j][i-1]][i-1];
            }
        }
        scanf("%d",&m);
        while(m--)
        {
            int l,r;
            int sum=1;
            scanf("%d%d",&l,&r);
            int v=p[r];
            for(int i=20;i>=0;i--)
            {
                if(anc[v][i]>=l){
                    sum+=(1<<i);
                    v=anc[v][i];
                }
            }
            printf("%d\n",sum/4);
        }
    }
    return 0;
}