Codeforces Round 934 (Div. 2) D. Non-Palindromic Substring

题目

思路：

#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
const int maxn = 1e6 + 5, inf = 1e9, maxm = 4e4 + 5;
const int N = 1e6;
// const int mod = 1e9 + 7;
// const int mod = 998244353;
const __int128 mod = 212370440130137957LL;
// int a[505][5005];
// bool vis[505][505];
// char s[505][505];
int a[maxn], b[maxn];
int vis[maxn];
string s;
int n, m;struct Node{int val, id;bool operator<(const Node &u)const{return val < u.val;}
};
// Node c[maxn];int ans[maxn];
int pre[maxn], suf[maxn];
int pw[maxn];
int base = 337;//long long ? maxn ?
// string s[maxn];
int t[maxn][26];//t[i][j]表示长度为i，字符全为char('a' + j)的字符串的哈希值int 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 Hash2(int l, int r){if(l > r){return 0;}return (suf[l] - (__int128)suf[r + 1] * pw[r - l + 1] % mod + mod) % mod;
}bool same(int l, int r){//判断字符串所有字符都相同for(int j = 0; j < 26; j++){if(Hash(l, r) == t[r - l + 1][j] && Hash2(l, r) == t[r - l + 1][j]){return 1;}}return 0;
}bool ispal(int l, int r){//判断回文return Hash(l, r) == Hash2(l, r);
}bool isalt(int l, int r){//字符串字符交替或字符全相同 转化为 前缀后缀分别去掉2个字符后相等return Hash(l, r - 2) == Hash(l + 2, r) && Hash2(l, r - 2) == Hash2(l + 2, r);
}void solve(){int res = 0;int q, k;int sum = 0;int mx = 0;cin >> n >> q;cin >> s;s = " " + s;for(int i = 1; i <= n; i++){pre[i] = ((__int128)pre[i - 1] * base % mod + s[i] - 'a') % mod;}suf[n + 1] = 0;for(int i = n; i >= 1; i--){suf[i] = ((__int128)suf[i + 1] * base % mod + s[i] - 'a') % mod;}while(q--){int l, r;cin >> l >> r;if(same(l, r)){cout << 0 << '\n';continue;}int len = r - l + 1;int sum = (len - 2) * (len + 1) / 2;//2 + 3 +...+ len - 1if(isalt(l, r) && len - 1 >= 3){//考虑字符交替的情况int cnt = (len - 1 - 3) / 2 + 1;sum -= cnt * (3 + 3 + 2 * (cnt - 1)) / 2;}if(!ispal(l, r)){//考虑k = len的情况sum += len;}cout << sum << '\n';}
}signed main(){ios::sync_with_stdio(0);cin.tie(0);pw[0] = 1;for(int i = 1; i <= N; i++){pw[i] = (__int128)pw[i - 1] * base % mod;}for(int j = 0; j < 26; j++){t[1][j] = j;}for(int i = 2; i <= N; i++){for(int j = 0; j < 26; j++){t[i][j] = ((__int128)t[i - 1][j] * base % mod + j) % mod;}}int T = 1;cin >> T;while (T--){solve();}return 0;
}