一个字符串的前缀是指包含该字符第一个字母的连续子串,例如:abcd的所有前缀为a, ab, abc, abcd。
给出一个字符串S,求其所有前缀中,字符长度与出现次数的乘积的最大值。
例如:S = “abababa” 所有的前缀如下:
“a”, 长度与出现次数的乘积 1 * 4 = 4,
“ab”,长度与出现次数的乘积 2 * 3 = 6,
“aba”, 长度与出现次数的乘积 3 * 3 = 9,
“abab”, 长度与出现次数的乘积 4 * 2 = 8,
“ababa”, 长度与出现次数的乘积 5 * 2 = 10,
“ababab”, 长度与出现次数的乘积 6 * 1 = 6,
“abababa”, 长度与出现次数的乘积 7 * 1 = 7.
其中"ababa"出现了2次,二者的乘积为10,是所有前缀中最大的。
输入
输入字符串S, (1 <= L <= 100000, L为字符串的长度),S中的所有字符均为小写英文字母。
输出
输出所有前缀中字符长度与出现次数的乘积的最大值。
输入样例
abababa
输出样例
10
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int sa[maxn], rak[maxn];
int tx[maxn], height[maxn];
char s[maxn];
int n, m, p;
int cnt[maxn];
struct node {
int x, y, id;
}a[maxn], b[maxn];
void rsort() {
for (int i = 1; i <= m; i++) tx[i] = 0;
for (int i = 1; i <= n; i++) tx[a[i].y]++;
for (int i = 1; i <= m; i++) tx[i] += tx[i - 1];
for (int i = 1; i <= n; i++) b[tx[a[i].y]--] = a[i];
for (int i = 1; i <= m; i++) tx[i] = 0;
for (int i = 1; i <= n; i++) tx[b[i].x]++;
for (int i = 1; i <= m; i++) tx[i] += tx[i - 1];
for (int i = n; i >= 1; i--) a[tx[b[i].x]--] = b[i];
}
void ssort() {
rsort();
p = 0;
for (int i = 1; i <= n; i++) {
if (a[i].x != a[i - 1].x || a[i].y != a[i - 1].y) ++p;
rak[a[i].id] = p;
}
for (int i = 1; i <= n; i++) {
a[i].x = rak[i];
a[i].id = sa[rak[i]] = i;
a[i].y = 0;
}
m = p;
}
void solve() {
m = 127;
for (int i = 1; i <= n; i++) {
a[i].x = a[i].y = s[i];
a[i].id = i;
}
ssort();
for (int j = 1; j <= n; j <<= 1) {
for (int i = 1; i + j <= n; i++) {
a[i].y = a[i + j].x;
}
ssort();
if (p == n) break;
}
}
void get_Height() {
int k = 0;
for (int i = 1; i <= n; i++) {
if (k) k--;
int j = sa[rak[i] - 1];
while (i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
height[rak[i]] = k;
}
}
int main() {
scanf("%s", s + 1);
n = strlen(s + 1);
solve(); get_Height();
int mm = 1e8;
for (int i = rak[1] + 1; i <= n; i++) {
mm = min(mm, height[i]);
cnt[mm]++;
}
mm = 1e8;
for (int i = rak[1]; i >= 1; i--) {
mm = min(mm, height[i]);
cnt[mm]++;
}
for (int i = n; i >= 1; i--) {
cnt[i] += cnt[i + 1];
}
long long ans = 0;
for (int i = 1; i <= n; i++) {
ans = max(ans, 1LL * i * (cnt[i] + 1));
}
printf("%lld\n", ans);
return 0;
}