前缀数组O(n^3)做法

前缀数组O(n^3)做法

s.substr（)的应用非常方便

令string s = "";

string sub1 = s.substr(); //只有一个数字5表示从下标为5开始一直到结尾：sub1 = "56789"

string sub2 = s.substr(, ); //从下标为5开始截取长度为3位：sub2 = "567"

#include<iostream>

#include<cstdio>

#include<queue>

#include<algorithm>

#include<cmath>

#include<ctime>

#include<set>

#include<map>

#include<stack>

#include<cstring>

#pragma GCC optimize(2)

#define inf 2147483647

#define ls rt<<1

#define rs rt<<1|1

#define lson ls,nl,mid,l,r

#define rson rs,mid+1,nr,l,r

#define N 1000010

#define For(i,a,b) for(int i=a;i<=b;++i)

#define p(a) putchar(a)

#define g() getchar()

using namespace std;

int len;

string s;

int ans[N];

void in(int &x){

    int y=;char c=g();x=;

    while(c<''||c>''){if(c=='-')y=-;c=g();}

    while(c<=''&&c>=''){ x=(x<<)+(x<<)+c-'';c=g();}

    x*=y;

}

void o(int x){

    if(x<){p('-');x=-x;}

    if(x>)o(x/);

    p(x%+'');

}

int main(){

    cin>>s;

    len=s.size();

    For(i,,len-)

        For(k,,i)

            if(s.substr(,k)==s.substr(i-k+,k))

                ans[i]=k;

    For(i,,len-)

        o(ans[i]),p(' ');

    return ;

}