题目链接：美好的每一天
经典题，这种字符串重排以后回文，优先考虑异或操作。考虑对一个字符串状态压缩，每个字符表示为 \(1<<(c-97)\)，然后将它们异或起来就可以得到一个字符串的状态压缩情况。
考虑每一位异或情况，如果为偶数个单个字符则为 \(0\)，奇数则为 \(1\)，注意可以重排，那么当且仅当 \(1\) 的数量 \(<=1\) 时为回文串。而区间异或，我们可以转化为前缀异或。
注意到前缀异或为 \(pre[r]-pre[l-1]\)，所以我们查询的时候还需要加入 \(l-1\) 处的点，保证个数不漏。
考虑单点加入和删除的贡献情况，第一种就是和自身异或为 \(0\) 的，第二种则是异或为某个 \(1\)，即 \(2\) 的次方，由于至多 \(26\) 种，又观察到 \(n \le 6 \times 10^4\)，那么枚举就行了。注意下，删除的时候先删除自身，因为删除会去掉和自身异或为 \(0\) 的情况，所以要优先排除自身和自身异或，增加则先加入和其他异或的情况，再加入自身。跑个普通莫队预处理出前缀异或数组即可，每次加入或者删除为前缀异或数组里的某个值。
参照代码
#include <bits/stdc++.h>
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
#define isPbdsFile
#ifdef isPbdsFile
#include <bits/extc++.h>
#else
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>
#endif
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
template <typename T>
int disc(T* a, int n)
{
return unique(a + 1, a + n + 1) - (a + 1);
}
template <typename T>
T lowBit(T x)
{
return x & -x;
}
template <typename T>
T Rand(T l, T r)
{
static mt19937 Rand(time(nullptr));
uniform_int_distribution<T> dis(l, r);
return dis(Rand);
}
template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
return (a % b + b) % b;
}
template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
a %= c;
T1 ans = 1;
for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
return modt(ans, c);
}
template <typename T>
void read(T& x)
{
x = 0;
T sign = 1;
char ch = getchar();
while (!isdigit(ch))
{
if (ch == '-')sign = -1;
ch = getchar();
}
while (isdigit(ch))
{
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
x *= sign;
}
template <typename T, typename... U>
void read(T& x, U&... y)
{
read(x);
read(y...);
}
template <typename T>
void write(T x)
{
if (typeid(x) == typeid(char))return;
if (x < 0)x = -x, putchar('-');
if (x > 9)write(x / 10);
putchar(x % 10 ^ 48);
}
template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
write(x), putchar(c);
write(c, y...);
}
template <typename T11, typename T22, typename T33>
struct T3
{
T11 one;
T22 tow;
T33 three;
bool operator<(const T3 other) const
{
if (one == other.one)
{
if (tow == other.tow)return three < other.three;
return tow < other.tow;
}
return one < other.one;
}
T3() { one = tow = three = 0; }
T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
{
}
};
template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
if (x < y)x = y;
}
template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
if (x > y)x = y;
}
constexpr int N = 6e4 + 10;
int pos[N];
struct Mo
{
int l, r, id;
bool operator<(const Mo& other) const
{
return pos[l] ^ pos[other.l] ? l < other.l : pos[l] & 1 ? r < other.r : r > other.r;
}
} node[N];
int pre[N];
ll ans;
constexpr int MX = 1 << 26 | 1;
int cnt[MX];
inline void add(const int curr)
{
ans += cnt[curr];
cnt[curr]++;
forn(i, 0, 25)ans += cnt[curr ^ 1 << i];
}
inline void del(const int curr)
{
cnt[curr]--;
ans -= cnt[curr];
forn(i, 0, 25)ans -= cnt[curr ^ 1 << i];
}
int n, q;
char c;
ll res[N];
inline void solve()
{
cin >> n >> q;
const int siz = sqrt(n);
forn(i, 1, n)cin >> c, pre[i] = 1 << c - 'a', pre[i] ^= pre[i - 1], pos[i] = (i - 1) / siz + 1;
forn(i, 1, q)
{
auto& [l,r,id] = node[i];
cin >> l >> r, id = i;
l--;
}
sortArr(node, q);
int l = 1, r = 0;
forn(i, 1, q)
{
auto [L,R,id] = node[i];
while (l < L)del(pre[l++]);
while (l > L)add(pre[--l]);
while (r < R)add(pre[++r]);
while (r > R)del(pre[r--]);
res[id] = ans;
}
forn(i, 1, q)cout << res[i] << endl;
}
signed int main()
{
// MyFile
Spider
//------------------------------------------------------
// clock_t start = clock();
int test = 1;
// read(test);
// cin >> test;
forn(i, 1, test)solve();
// while (cin >> n, n)solve();
// while (cin >> test)solve();
// clock_t end = clock();
// cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}