动态规划基础题(第3题)：AC累积和

N Q
S
l1 r1
:
lQ rQ

8 3
ACACTACG
3 7
2 3
1 8

2
0
3

void coder_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, q;
    cin >> n >> q;
    string s;
    cin >> s;
    vector<int> vi(n + 1);
    vi[0] = 0;
    long long C, P;
    for(int i = 1; i <= n; i++) {
        if(s[i - 1] == 'A' && s[i] == 'C') {
            vi[i + 1] = vi[i] + 1;
            continue;
        }
        vi[i + 1] = vi[i];
    }
    int L, R;
    for(int i = 0; i < q; i++) {
        cin >> L >> R;
        cout << vi[R] - vi[L]<< endl;
    }
}

int N, Q; string S;
int ok[101010], sm[101010];
//---------------------------------------------------------------------------------------------------
void _main() {
	cin >> N >> Q >> S;
	rep(i, 0, N - 1) if (S[i] == 'A' and S[i + 1] == 'C') ok[i] = 1;
	sm[0] = ok[0];
	rep(i, 1, N) sm[i] = sm[i - 1] + ok[i];
 
	rep(q, 0, Q) {
		int L, R; cin >> L >> R; L--; R--;
 
		int ans = sm[R-1];
		if (L) ans -= sm[L - 1];
		
		printf("%d\n", ans);
	}
}

#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
#define FI first
#define SE second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
 
 
int main()
{
    lint N;
    cin >> N;
    lint Q;
    cin >> Q;
    string S;
    cin >> S;
    vector<lint> l(Q);
    vector<lint> r(Q);
    REP(i, Q){ cin >> l[i]; cin >> r[i]; }
 
    vector<int> s(N + 1);
    REP(i, N - 1) s[i + 1] = s[i] + (S[i] == 'A' and S[i + 1] == 'C');
 
    REP(i, Q)
    {
        cout << s[r[i] - 1] - s[l[i] - 1] << endl;
    }
    return 0;
}