As Easy As Possible

题意：给定一个字符串，有多段查询，问 (l,r) 中最多出现了多少个 easy

倍增

先预处理，对于每一个e往后跳 2**0 步是最接近它的a，每个a往后跳 2**0步是最接近它的s，以此类推。y后面要接e，于是就用倍增，利用f[i][j]表示第i位往后跳2**j步所到达的位置。然后对于每次查询，先找到第一个大于等于l的 e，然后看最多能往后面跳多少步，结果除以4就是答案。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
const int maxn = 200000;
char s[maxn];
int se[maxn], sa[maxn], ss[maxn], sy[maxn];
int f[maxn][100], x, y, l, r, mid, t, step, n, len;
int main() {
scanf("%s", s);
len = strlen(s);
for (int i = len; i >= 1; i--) s[i] = s[i-1];
for (int i = 1; i <= len; i++) {
if (s[i] == 'e') {
se[++se[0]] = i;
for (int j = sy[0]; j >= 1; j--) {
if (f[sy[j]][0] != 0) break;
f[sy[j]][0] = i;
}
} else if (s[i] == 'a') {
sa[++sa[0]] = i;
for (int j = se[0]; j >= 1; j--) {
if (f[se[j]][0] != 0) break;
f[se[j]][0] = i;
}
} else if (s[i] == 's') {
ss[++ss[0]] = i;
for (int j = sa[0]; j >= 1; j--) {
if (f[sa[j]][0] != 0) break;
f[sa[j]][0] = i;
}
} else if (s[i] == 'y') {
sy[++sy[0]] = i;
for (int j = ss[0]; j >= 1; j--) {
if (f[ss[j]][0] != 0) break;
f[ss[j]][0] = i;
}
}
}
for (int i = 1; i <= len; i++) if (f[i][0] == 0) f[i][0] = maxn;
for (int j = 0; 1<<j <= len; j++)
for (int i = 1; i <= len; i++) {
if (f[i][j] == maxn) f[i][j+1] = maxn;
else f[i][j+1] = f[f[i][j]][j];
}
scanf("%d", &n);
while (n--) {
scanf("%d%d", &x, &y);
l = 1; r = se[0]; t = 0;
while (l <= r) {
mid = (l+r)/2;
if (se[mid] >= x) {
t = se[mid];
r = mid-1;
} else l = mid+1;
}
if (t == 0 || t > y) {
printf("0\n");
continue;
}
step = 0;
while (t <= y) {
if (f[t][0] > y) break;
for (int j = 1; j <= len+1; j++) if (f[t][j] > y) {
step += 1<<(j-1);
t = f[t][j-1];
break;
}
}
printf("%d\n", (step+1)/4);
}
}