*[codility]GenomicRangeQuery

http://codility.com/demo/take-sample-test/genomicrangequery

这题有点意思。
一开始以为是RMQ或者线段树，但这样要O(n*logn)。考虑到只有四种字符，可以用数组记录每个字符i之前出现过几次。
二，查询区间是闭区间，所以要处理off by one的问题。

// you can also use includes, for example:
// #include <algorithm>
vector<int> solution(string &S, vector<int> &P, vector<int> &Q) {
    // write your code in C++98
    int len = S.length();
    vector<int> A(len+1);
    vector<int> C(len+1);
    vector<int> G(len+1);
    vector<int> T(len+1);
    A[0] = 0;
    C[0] = 0;
    G[0] = 0;
    T[0] = 0;
    for (int i = 1; i <= len; i++) {
        A[i] = A[i-1];
        C[i] = C[i-1];
        G[i] = G[i-1];
        T[i] = T[i-1];
        if (S[i-1] == 'A') {
            A[i]++;
        }
        else if (S[i-1] == 'C') {
            C[i]++;
        }
        else if (S[i-1] == 'G') {
            G[i]++;
        }
        else if (S[i-1] == 'T') {
            T[i]++;
        }
    }
    vector<int> ans;
    for (int i = 0; i < P.size(); i++) {
        int p = P[i];
        int q = Q[i] + 1;
        if (A[q] - A[p] > 0) ans.push_back(1);
        else if (C[q] - C[p] > 0) ans.push_back(2);
        else if (G[q] - G[p] > 0) ans.push_back(3);
        else if (T[q] - T[p] > 0) ans.push_back(4);
    }
    return ans;
}