ACdream 1116 扩展KMP 线段树区间累加 Fib数列的矩阵加速

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Gao the string!

[align=center]Time Limit: 2000/1000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)[/align]
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Problem

Problem Description

give you a string, please output the result of the following function mod 1000000007



n is the length of the string

f() is the function of fibonacci, f(0) = 0, f(1) = 1...

a[i] is the total number of times any prefix appear in the suffix s[i....n-1].

(the prefix means s[0...i] )

Input

multiple test cases.

each test case will give you a string consists of lowercase letters, the length of which is no more than 100000.

Output

ouput a number.

Sample Input

aa

Sample Output

3

Hint

样例解释如下：

对于

aa这个后缀，前缀a出现了两次，前缀aa出现了一次，总共三次

对于a这个后缀，前缀a出现了一次

所以答案是f(3) + f(1)

Source

wuyiqi

题意：给出一个长度小于10W的字符串。用a[I]表示所以前缀在后缀s[I..n]中出现的次数。求

的值 mod 1000000007，f()为fib数列。

思路：求出字符串每一位的exnext值。(exnext的第一位在这道题强行定位s的长度)。用线段树维护a[I]的值，假设next[I]=k。则在后缀1到后缀I中，都出现了前缀1到前缀k。所以将区间[1,I]都加上exnext[I]。最后将a[I]都导出导到一个数组里面（不导出也可以）方便后面的计算。由于a[ii[]很大，可能超过int。所以求f(a[I])必须使用矩阵加速。

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#define LL long long
#define maxn 100005
#define MOD 1000000007
using namespace std;
int exnext[maxn];
void getexnext(char s[])
{
int len = strlen(s);
exnext[0] = 0;
int k, j, i;
i = j = 0;
for (k = 1; k < len; k++)
{
if (k + exnext[k - i] > j)
{
i = k;
while (j < len&&s[j - i + 1] == s[j + 1]) j++;
exnext[k] = j - k + 1;
if (j < k) j = k;
}
else exnext[k] = k + exnext[k - i] - 1 < len ? exnext[k - i] : len - k + 1;
}
}
struct node
{
int l,r;
LL add;
} tree[maxn<<2];
void build(int id,int l,int r)
{
tree[id].l=l;
tree[id].r=r;
tree[id].add=0;
if(l==r) return;
int mid=(l+r)>>1;
build(id<<1,l,mid);
build(id<<1|1,mid+1,r);
}
void op(int id,int l,int r,int add)
{
if(l<=tree[id].l&&tree[id].r<=r) tree[id].add+=add;
else
{
int mid=(tree[id].l+tree[id].r)>>1;
if(l<=mid) op(id<<1,l,r,add);
if(mid<r) op(id<<1|1,l,r,add);
}
}
void pushdown(int id)
{
tree[id<<1].add+=tree[id].add;
tree[id<<1|1].add+=tree[id].add;
}
LL Ans[maxn];
void out(int id)
{
if(tree[id].l==tree[id].r) Ans[tree[id].l]=tree[id].add;
else
{
pushdown(id);
out(id<<1);
out(id<<1|1);
}
}
struct matrix
{
LL a[2][2];
};
matrix mul(matrix& x,matrix& y)
{
matrix ans;
ans.a[0][0]=x.a[0][0]*y.a[0][0]+x.a[0][1]*y.a[1][0];
ans.a[0][0]%=MOD;
ans.a[0][1]=x.a[0][0]*y.a[0][1]+x.a[0][1]*y.a[1][1];
ans.a[0][1]%=MOD;
ans.a[1][0]=x.a[1][0]*y.a[0][0]+x.a[1][1]*y.a[1][0];
ans.a[1][0]%=MOD;
ans.a[1][1]=x.a[1][0]*y.a[0][1]+x.a[1][1]*y.a[1][1];
ans.a[1][1]%=MOD;
return ans;
}
matrix power(matrix x,long e)
{
matrix ans,tmp;
if(e==0)
{
ans.a[0][0]=1;
ans.a[0][1]=0;
ans.a[1][0]=0;
ans.a[1][1]=1;
return ans;
}
if( e==1 )
return x;
tmp=power(x,e>>1);
ans=mul(tmp,tmp);
if( e&1 )
ans=mul(ans,x);
return ans;
}
LL fic(int n)
{
matrix result = {{1,0,0,1}};
matrix base = {{1,1,1,0}};
if(n&1) result = base;
n>>=1;
while(n)
{
base=mul(base, base);
if(n&1)
{
result=mul(result, base);
}
n>>=1;
}
return result.a[0][1];
}
char s[maxn];
int main()
{
while(~scanf("%s",s))
{
getexnext(s);
int len=strlen(s);
LL ans=0;
exnext[0]=len;
build(1,0,len-1);
for(int i=0;i<len;i++) op(1,0,i,exnext[i]);
out(1);
for(int i=0;i<len;i++) ans=(ans+fic(Ans[i]))%MOD;
printf("%lld\n",ans);
}
return 0;
}