CS Academy Prefix Matches

题目链接：https://csacademy.com/contest/archive/task/prefix-matches/

题目大意：给出字符串S，现在规定数列Ai为S中从i开始的字符串的前缀的最大长度，并且该前缀也出现在S的前缀中；数列Bi为S中以i为结尾的字符串的前缀最大长度，该前缀至少从i=2开始，并且该前缀也出现在S的所有前缀中。我们不知道S，但给出A，求B。

解题思路：可以发现，A中出现的每一个都在B中有对应关系。因此针对每个Ai，B中从i开始知道i+Ai的每个位置都对应一个前缀，长度从1递增。又由于长度递增，因此如果Ai，Aj对应B的区间出现重复，那么只需要考虑Ai在重复区间产生的影响即可。

代码：

const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
int a[maxn], b[maxn];
int n;

void solve(){
    memset(b, 0, sizeof(b));
    int i = 2, j = 2;
    while(i <= n){
        if(a[i] != 0){
            if(j <= i + a[i] - 1){
                for(int k = j - i + 1; k <= a[i]; k++){
                    b[j] = k; j++;
                }
            }
        }
        if(j == i) j++;
        i++;
        
    }
    
    for(int i = 2; i <= n; i++) {
        printf("%d", b[i]);
        if(i != n) printf(" ");
        else puts("");
    }
}
int main(){
    scanf("%d", &n);
    memset(a, 0, sizeof(a));
    for(int i = 2; i <= n; i++) scanf("%d", &a[i]);
    solve();
} 

题目：

Prefix Matches

Time limit: 1000 ms
Memory limit: 128 MB

 

For a string SS of size NN, we say A_iA​i​​ is the size of the longest prefix that can be found again in SS starting at index ii. A_1A​1​​ is not defined.

On the other hand, B_iB​i​​ is the size of the longest prefix that can be found again in SS ending at index ii. B_iB​i​​ is strictly less than ii, B_1B​1​​ is not defined.

For example, if S=ababacabaS=ababacaba, A=[-,0, 3, 0,1,0,3,0,1 ]A=[−,0,3,0,1,0,3,0,1] and B=[-,0,1,2, 3,0,1,2,3]B=[−,0,1,2,3,0,1,2,3]

You don't know SS, but given AA, find BB.

Standard input

The first line contains a single integer NN.

The second line contains N-1N−1 integers. The i^{th}i​th​​ number is A_{i+1}A​i+1​​.

Standard output

Print N-1N−1 value on the first line. The i^{th}i​th​​ number is B_{i+1}B​i+1​​.

Constraints and notes

• 2 leq N leq 10^52≤N≤10​5​​ 
• It is always possible to generate a string SS corresponding to AA using an alphabet of size at most 10^510​5​​.
• It is guaranteed the answer is unique.

Input	Output	Explanation
6 2 1 0 2 1	1 2 0 1 2	A possible string is: aaabaaaaabaa
5 4 3 2 1	1 2 3 4	aaaaaaaaaa
6 0 3 0 1 0	0 1 2 3 0	ababacababac
11 0 1 0 3 0 3 0 2 0 0	0 1 0 1 2 3 2 3 2 0	abacabababcabacabababc