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 - 1
        if(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;
}