2016多校训练Contest5: 1005 Interesting hdu5785

Problem Description

Alice get a string S. She thinks palindrome string is interesting. Now she wanna know how many three tuple (i,j,k) satisfy 1≤i≤j<k≤length(S),
S[i..j] and S[j+1..k] are all palindrome strings. It's easy for her. She wants to know the sum of i*k of all required three tuples. She can't solve it. So can you help her? The answer may be very large, please output the answer mod 1000000007.

A palindrome string is a string that is same when the string is read from left to right as when the string is read from right to left.

Input

The input contains multiple test cases.

Each test case contains one string. The length of string is between 1 and 1000000. String only contains lowercase letter.

Output

For each test case output the answer mod 1000000007.

Sample Input

aaa
abc

Sample Output

14
8

枚举j

马拉车处理每个点延伸的最长回文串，把可以延伸的部分加上一个等差数列

向左和向右延伸分开统计

这里可以打标记二维前缀和，考场上写了个树状数组统计结果T得飞起来了

然后直接累加s1[j]*s2[j+1]即可

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1000110;
char str[maxn];//原字符串
char tmp[maxn<<1];//转换后的字符串
int Len[maxn<<1];
int lx;
long long mod=1000000007;
//转换原始串
inline int INIT(char *st,int len)
{
int i;
tmp[0]='@';//字符串开头增加一个特殊字符，防止越界
for(i=1;i<=2*len;i+=2)
{
tmp[i]='#';
tmp[i+1]=st[i/2];
}
tmp[2*len+1]='#';
tmp[2*len+2]='$';//字符串结尾加一个字符，防止越界
tmp[2*len+3]=0;
return 2*len+1;//返回转换字符串的长度
}
//Manacher算法计算过程
inline void MANACHER(char *st,int len)
{
memset(Len,0,sizeof(Len));
int mx=0,ans=0,po=0;//mx即为当前计算回文串最右边字符的最大值
for(int i=1;i<=len;i++)
{
if(mx>i)
Len[i]=min(mx-i,Len[2*po-i]);//在Len[j]和mx-i中取个小
else
Len[i]=1;//如果i>=mx，要从头开始匹配
while(st[i-Len[i]]==st[i+Len[i]])
Len[i]++;
if(Len[i]+i>mx)//若新计算的回文串右端点位置大于mx，要更新po和mx的值
{
mx=Len[i]+i;
po=i;
}
}  //返回Len[i]中的最大值-1即为原串的最长回文子串额长度
}
struct tree
{
long long x1,x2,x3;
}tr[2][1001001];
inline int lowbit(int x)
{
return x&(-x);
}
struct save
{
long long x,y;
}sx1[1000001],sx2[1000001];
inline void add(int d,int l,long long x1,long long x2,long long x3)
{
if(l<=0)
l=1;
int i;
for(i=l;i<=lx;i+=lowbit(i))
{
tr[d][i].x1+=x1;
tr[d][i].x2+=x2;
tr[d][i].x3+=x3;
}
}
inline long long sum(int d,int x)
{
if(x>lx)
x=lx;
int i;
long long x1=0,x2=0,x3=0;
for(i=x;i>=1;i-=lowbit(i))
{
x1+=tr[d][i].x1;
x2+=tr[d][i].x2;
x3+=tr[d][i].x3;
}
return (long long)2*x2+x3-(long long)x*x1;
}
int main()
{
freopen("1005.in","r",stdin);
freopen("1005.ans","w",stdout);
while(scanf("%s",str)!=EOF)
{
lx=strlen(str);
int i;
/*	for(i=0;i<=lx-1;i++)
str[i]=ss[i];*/
int l=INIT(str,lx);
MANACHER(tmp,l);
//	memset(tr,0,sizeof(tr));
//for(i=1;i<=lx;i++)
//	tr[0][i].x1=tr[0][i].x2=tr[0][i].x3=tr[1][i].x1=tr[1][i].x2=tr[1][i].x3=0;
for(i=1;i<=lx*2;i++)
Len[i]--;
/*	for(i=2;i<=lx*2;i++)
{
if(tmp[i]>='a'&&tmp[i]<='z')
{
add(0,(i+1)/2,1,i/2,0);
add(0,(i+Len[i])/2+1,-1,-i/2,0);
}
else
{
add(0,(i+1)/2,1,i/2,1);
add(0,(i+Len[i])/2+1,-1,-i/2,-1);
}
}
for(i=2;i<=lx*2;i++)
{
if(tmp[i]>='a'&&tmp[i]<='z')
{
add(1,(i-Len[i]+1)/2,1,i/2,0);
add(1,i/2+1,-1,-i/2,0);
}
else
{
add(1,(i-Len[i]+1)/2,1,i/2,1);
add(1,i/2+1,-1,-i/2,-1);
}
}*/
memset(sx1,0,sizeof(sx1));
memset(sx2,0,sizeof(sx2));
for(i=2;i<=lx*2;i++)
{
if(tmp[i]>='a'&&tmp[i]<='z')
{
int l=(i+1)/2,r=(i+Len[i])/2;
if(l<=0) l==1;
sx1[l].x+=l;
sx1[l].y-=1;
sx1[r+1].x-=l*2-r;
sx1[r].y+=1;
}
else
{
int l=(i+1)/2,r=(i+Len[i])/2;
if(l>r)
continue;
if(l<=0) l==1;
sx1[l].x+=(i+1)/2-1;
sx1[l].y-=1;
sx1[r+1].x-=l*2-1-r;
sx1[r].y+=1;
}
}
for(i=2;i<=lx*2;i++)
{
if(tmp[i]>='a'&&tmp[i]<='z')
{
int l=(i-Len[i]+1)/2,r=i/2;
if(l<=0) l==1;
sx2[l].x+=r*2-l;
sx2[l].y-=1;
sx2[r+1].x-=r;
sx2[r].y+=1;
}
else
{
int l=(i-Len[i]+1)/2,r=i/2;
if(l<=0) l==1;
sx2[l].x+=r*2+1-l;
sx2[l].y-=1;
sx2[r+1].x-=r+1;
sx2[r].y+=1;
}
}
long long ans=0;
long long s1=0,s2=sx2[1].x;
long long dx1=0,dx2=sx2[1].y;
for(i=1;i<=lx-1;i++)
{
s1+=sx1[i].x+dx1;
dx1+=sx1[i].y;
s2+=dx2+sx2[i+1].x;
dx2+=sx2[i+1].y;
ans=(ans+(s1%mod)*(s2%mod)%mod)%mod;
}
printf("%I64d\n",ans);
}
return 0;
}