# WC analog (1.2) T1 string

strand

Background:

1.2 WC analog T1

Analysis: conclusion analysis

It seems that the answer can only be - 1,1,2, and then it directly judges 1, then n2 enumerates, delete where to delete, and then 60 points

If s is not a palindrome string, then the answer is obviously 1, then for 1 < I < n, if there i s an I that satisfies s[1..i] and s[i +I n 1..n], neither of them i s a palindrome string, so the answer should be 2. If any I satisfies at least one of them is a palindrome string, after discussion, it can be found that if s[1] ==s[2], then s should be in the shape of aaaaaaaa (all the same), or aaabaaa (the length is odd, only the middle one is different), if s[1] ! = s[2], then s shape is like abababa (palindrome string with odd length and alternating two letters). All three cases will be unsolved, so we only need to judge 1 (not palindrome string), - 1 (above three cases) and other output 2······

Source:

```/*
created by scarlyw
*/
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cctype>
#include <vector>
#include <set>
#include <queue>
#include <ctime>
#include <bitset>

static const int IN_LEN = 1024 * 1024;
static char buf[IN_LEN], *s, *t;
if (s == t) {
t = (s = buf) + fread(buf, 1, IN_LEN, stdin);
if (s == t) return -1;
}
return *s++;
}

///*
template<class T>
inline void R(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return ;
if (c == '-') iosig = true;
}
for (x = 0; isdigit(c); c = read())
x = ((x << 2) + x << 1) + (c ^ '0');
if (iosig) x = -x;
}
//*/

const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN], *oh = obuf;
inline void write_char(char c) {
if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;
*oh++ = c;
}

template<class T>
inline void W(T x) {
static int buf[30], cnt;
if (x == 0) write_char('0');
else {
if (x < 0) write_char('-'), x = -x;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
while (cnt) write_char(buf[cnt--]);
}
}

inline void flush() {
fwrite(obuf, 1, oh - obuf, stdout);
}

/*
template<class T>
inline void R(T &x) {
static char c;
static bool iosig;
for (c = getchar(), iosig = false; !isdigit(c); c = getchar())
if (c == '-') iosig = true;
for (x = 0; isdigit(c); c = getchar())
x = ((x << 2) + x << 1) + (c ^ '0');
if (iosig) x = -x;
}
//*/

const int MAXN = 100000 + 10;

int n, t;
char s[MAXN];

inline void solve() {
scanf("%d%s", &n, s);
bool flag = false ;
for (int i = 0; i < n; ++i)
if (s[i] != s[n - i - 1]) {
flag = true;
break ;
}
if (flag) {
W(1), write_char('\n');
return ;
}
if (n & 1) {
bool flag;
flag = true ;
for (int i = 0; i < n; ++i)
if (s[i] != s[i % 2]) {
flag = false;
break ;
}
if (flag) {
W(-1), write_char('\n');
return ;
}
flag = true ;
for (int i = 0; i < n; ++i)
if (i != n / 2 && s[i] != s[0]) {
flag = false;
break ;
}
if (flag) W(-1), write_char('\n');
else W(2), write_char('\n');
} else {
bool flag = true ;
for (int i = 0; i < n; ++i)
if (s[i] != s[0]) {
flag = false;
break ;
}
if (flag) W(-1), write_char('\n');
else W(2), write_char('\n');
}
}

int main() {
freopen("string.in", "r", stdin);
freopen("string.out", "w", stdout);
scanf("%d", &t);
while (t--) solve();
flush();
return 0;
}```

Posted by Stryker250 on Fri, 01 May 2020 02:50:45 -0700