As Easy As Possible

题目链接

题意：一个只含e, a, s, y的字符串，问[l, r]内有多少个easy序列？

题解：倍增。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5+10;
char s[N];
int id(char c){
    if(c == 'e') return 0;
    if(c == 'a') return 1;
    if(c == 's') return 2;
    return 3;
}
int Log[N];
int f[17][N], nex[N][4];
int main(){
    for(int i = 2; i < N; i++) Log[i] = Log[i>>1]+1;
    scanf("%s", s+1);
    int len = strlen(s+1);
    int last[4];
    last[0] = last[1] = last[2] = last[3] = len+1;
    for(int i = len; i; i--){
        int k = id(s[i]);
        last[k] = i;
        for(int j = 0; j < 4; j++)
            nex[i][j] = last[j];
    }
    //nex[i][k]: [i, len]中第一个k出现的位置
    for(int i = 1; i <= len; i++){
        f[0][i] = nex[i][0];
        for(int j = 1; j < 4&&f[0][i] <= len; j++)
            f[0][i] = nex[f[0][i]][j];
    }
    //f[0][i]: [i, len]中第一个出现easy的结尾处
    for(int j = 1; j <= Log[len]; j++)
        for(int i = 1; i <= len; i++){
            int pos = f[j-1][i];
            f[j][i] = pos > len? pos : f[j-1][pos];
        }

    int m, l, r;
    scanf("%d", &m);
    while(m--){
        scanf("%d%d", &l, &r);
        int ans = 0;
        for(int i = Log[len]; i >= 0&&l < r; i--){
            if(f[i][l] > r) continue ;
            ans += 1<<i;
            l = f[i][l]+1;
        }
        printf("%d\n", ans);
    }
    return 0;
}