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Description
Input
一行，一个字符串S
Output
一行，一个整数，表示所求值
Sample Input
cacao
Sample Output
54
HINT
2<=N<=500000,S由小写英文字母组成
题解
预处理height
预处理一个ST表
考虑一个位置的height能做的贡献
显然是这个位置往前往后连续的一段
二分段
但是可能会出现一个问题
height成了
…3…3…
中间都是比3大的数
那么这样如果两边都允许相等 会出现重复的贡献被计算
强行让左边不等于 右边允许等于就可以了.
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*/
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#define LL long long
using namespace std;
char ch[500005];
int Rank[500005],Rsort[500005],a[500005],sa1[500005],sa2[500005],tt[500005];
void get_sa(int len,int m)
{
memcpy(Rank,a,sizeof(Rank));
memset(Rsort,0,sizeof(Rsort));
for(int i=1;i<=len;i++)Rsort[Rank[i]]++;
for(int i=1;i<=m;i++)Rsort[i]+=Rsort[i-1];
for(int i=len;i>=1;i--)sa1[Rsort[Rank[i]]--]=i;
int ln=1,p=0;
while(p<len)
{
int k=0;
for(int i=len-ln+1;i<=len;i++)sa2[++k]=i;
for(int i=1;i<=len;i++)if(sa1[i]-ln>0)sa2[++k]=sa1[i]-ln;
memset(Rsort,0,sizeof(Rsort));
for(int i=1;i<=len;i++)Rsort[Rank[i]]++;
for(int i=1;i<=m;i++)Rsort[i]+=Rsort[i-1];
for(int i=len;i>=1;i--)sa1[Rsort[Rank[sa2[i]]]--]=sa2[i];
memcpy(tt,Rank,sizeof(tt));
Rank[sa1[1]]=1;p=1;
for(int i=2;i<=len;i++)
{
if(tt[sa1[i]]!=tt[sa1[i-1]]||tt[sa1[i]+ln]!=tt[sa1[i-1]+ln])p++;
Rank[sa1[i]]=p;
}
ln*=2;m=p;
}
}
int height[500005];
void get_he(int len)
{
int j,k=0;
for(int i=1;i<=len;i++)
{
j=sa1[Rank[i]-1];
if(k)k--;
while(ch[i+k]==ch[j+k])k++;
height[Rank[i]]=k;
}
}
int mn[500005][25],bin[25],Log[500005];
int getmn(int l,int r)
{
int K=Log[r-l+1];
return min(mn[l][K],mn[r-bin[K]+1][K]);
}
int main()
{
// freopen("a.in","r",stdin);
// freopen("a.out","w",stdout);
Log[1]=0;for(int i=2;i<=500000;i++)Log[i]=Log[i>>1]+1;
bin[0]=1;for(int i=1;i<=20;i++)bin[i]=bin[i-1]<<1;
scanf("%s",ch+1);int len=strlen(ch+1);
for(int i=1;i<=len;i++)a[i]=ch[i]-'a'+1;
get_sa(len,26);
get_he(len);
LL ans=((LL)(1+len))*((LL)(len))/2*((LL)len-1);
//height[1]=999999999;
for(int i=1;i<=len;i++)mn[i][0]=height[i];
for(int i=1;bin[i]<=len;i++)
for(int j=1;j+bin[i]-1<=len;j++)
mn[j][i]=min(mn[j][i-1],mn[j+bin[i-1]][i-1]);
for(int i=2;i<=len;i++)
{
int l=2,r=i,fa=i,fb=i;
while(l<=r)
{
int mid=(l+r)/2;
if(getmn(mid,i-1)>height[i])fa=mid,r=mid-1;
else l=mid+1;
}//左边不越界
fa--;
l=i,r=len;
while(l<=r)
{
int mid=(l+r)/2;
if(getmn(i,mid)>=height[i])fb=mid,l=mid+1;
else r=mid-1;
}//右边越界
ans-=2LL*(LL)height[i]*((LL)(i-1)-fa+1)*((LL)fb-i+1);
}
printf("%lld\n",ans);
return 0;
}