【开源社区openEuler实践】 E. No Palindromes

题目

题解

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
#define ll long long
#define pii pair<int, int>
#define ld long double
const int maxn = 1e6 + 5, inf = 1e18, maxm = 4e4 + 5, base = 337;
const int N = 2e6;
// const int mod = 1e9 + 7;
// const int mod = 998244353;
const __int128 mod = 212370440130137957LL;
int n, m;

int pre[maxn], pw[maxn];
int suf[maxn];
int pre_hash(int l, int r){
    if(l > r) return 0;
    return (pre[r] - (__int128)pre[l - 1] * pw[r - l + 1] % mod + mod) % mod;
}
int suf_hash(int l, int r){
    if(l > r) return 0;
    return (suf[l] - (__int128)suf[r + 1] * pw[r - l + 1] % mod + mod) % mod;
}
bool check(int l, int r){
    return pre_hash(l, r) == suf_hash(l, r);
}
//long long ? maxn ? n? m?
void solve(){
    ll res = 0;
    int k;
    // cin >> n;
	// vector<int> a(n + 1);
	// for(int i = 1; i <= n; i++){
	// 	cin >> a[i];
	// }
    string s;
    cin >> s;
    n = s.size();
    s = ' ' + s;
    for(int i = 1; i <= n; i++){
        pre[i] = ((__int128)pre[i - 1] * base + s[i]) % mod;
    }
    suf[n + 1] = 0;
    for(int i = n; i >= 1; i--){
        suf[i] = ((__int128)suf[i + 1] * base + s[i]) % mod;
    }
    if(!check(1, n)){
        cout << "Yes\n";
        cout << 1 << '\n' << s.substr(1) << '\n';
        return;
    }
    for(int i = 1; i < n; i++){
        if(!check(1, i) && !check(i + 1, n)){
            cout << "Yes\n";
            cout << 2 << '\n';
            cout << s.substr(1, i) << ' ' << s.substr(i + 1, n - (i + 1) + 1) << '\n';
            return;
        }
    }
    cout << "No\n";
    // cout << res << '\n';
}
    
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(9);
    
    pw[0] = 1;
    for(int i = 1; i <= N; i++){
        pw[i] = (__int128)pw[i - 1] * base % mod;
    }
    int T = 1;
    cin >> T;
    while (T--){
        solve();
    }
    return 0;
}