The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm:
Step1. Connect the father’s name and the mother’s name, to a new string S.
Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S).
Example: Father=‘ala’, Mother=‘la’, we have S = ‘ala’+‘la’ = ‘alala’. Potential prefix-suffix strings of S are {‘a’, ‘ala’, ‘alala’}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:)
Input
The input contains a number of test cases. Each test case occupies a single line that contains the string S described above.
Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.
Output
For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby’s name.
Sample Input
ababcababababcabab
aaaaa
Sample Output
2 4 9 18
1 2 3 4 5
输入一个字符串s，求出既是s的前缀又是s的后缀的字符串长度，并从小到大输出
注意s的长度n一定满足条件
已知s的长度n是最大的长度，Next[n]即为s第二大的长度，Next[Next[n]]为第三大的长度…以此类推，直到遍历到的Next值为0，说明已经没有更小的长度
可以将这些数字放入一个栈内，再依次弹出即可
#include <cstdio>
#include <iostream>
#include <cstring>
#include <stack>
using namespace std;
const int N = 1e6 + 5;
char a[N], b[N];
int Next[N], f[N]; // Next是字符串 a 和自己匹配，f是字符串 a 和 b 匹配
int n, m;
// 求解 Next 数组
void get_next()
{
Next[1] = 0;
for (int i = 2, j = 0; i <= n; i++) {
while (j > 0 && a[i] != a[j + 1]) j = Next[j];
if (a[i] == a[j + 1]) j++;
Next[i] = j;
}
}
int main(void)
{
stack<int> stk;
while (scanf("%s", a + 1) != EOF) {
n = strlen(a + 1);
get_next();
stk.push(n);
while (Next[n] != 0) {
stk.push(Next[n]);
n = Next[n];
}
while (stk.size()) {
cout << stk.top() << " ";
stk.pop();
}
cout << endl;
}
return 0;
}