*[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;

}