题目
给定一个只包含小写字母的字符串SS,
请你求出 SS 的所有出现次数不为 11 的子串的出现次数乘上该子串长度的最大值。
输入格式
一行一个仅包含小写字母的字符串SS
输出格式
一个整数，为 所求答案
输入样例
abab
输出样例
4
提示
对于\(10\%10%\)的数据，|S|<=1000∣S∣<=1000
对于\(100\%100%\)的数据，|S|<=\(10^6\)∣S∣<=\(10^6\)
题解
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u]; k; k = ed[k].nxt)
using namespace std;
const int maxn = 2000005,maxm = 100005,INF = 1000000000;
inline int RD(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57) {if (c == ‘-‘) flag = -1; c = getchar();}
while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - ‘0‘; c = getchar();}
return out * flag;
}
int ch[maxn][26],pre[maxn],step[maxn],n,cnt,last;
int b[maxn],sz[maxn],a[maxn];
LL ans = 0;
char s[maxn];
void ins(int u){
int p = last,np = ++cnt;
last = np; step[np] = step[p] + 1;
while (p && !ch[p][u]) ch[p][u] = np,p = pre[p];
if (!p) pre[np] = 1;
else {
int q = ch[p][u];
if (step[q] == step[p] + 1) pre[np] = q;
else {
int nq = ++cnt; step[nq] = step[p] + 1;
for (int i = 0; i < 26; i++) ch[nq][i] = ch[q][i];
pre[nq] = pre[q]; pre[q] = pre[np] = nq;
while (ch[p][u] == q) ch[p][u] = nq,p = pre[p];
}
}
sz[np] = 1;
}
void solve(){
REP(i,cnt) b[step[i]]++;
REP(i,cnt) b[i] += b[i - 1];
REP(i,cnt) a[b[step[i]]--] = i;
for (int i = cnt; i; i--){
sz[pre[a[i]]] += sz[a[i]];
if (sz[a[i]] > 1) ans = max(ans,1ll * step[a[i]] * sz[a[i]]);
}
}
int main(){
scanf("%s",s + 1);
cnt = last = 1; n = strlen(s + 1);
for (int i = 1; i <= n; i++) ins(s[i] - ‘a‘);
solve();
printf("%lld",ans);
return 0;
}