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 = "0123456789";

string sub1 = s.substr(5); //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(5, 3); //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=1;char c=g();x=0;
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=g();}
    while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=g();}
    x*=y;
}
void o(int x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

int main(){
    cin>>s;
    len=s.size();
    For(i,1,len-1)
        For(k,1,i)
            if(s.substr(0,k)==s.substr(i-k+1,k))
                ans[i]=k;
    For(i,0,len-1)
        o(ans[i]),p(' ');
    return 0;
}