#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#define ll long long
int n, m;
std::vector<int> k, p;
inline int mpow(int x, int k) {
    int ans = 1;
    for (; k; k >>= 1, x = x * x) {
        if (k & 1)
            ans = ans * x;
    }
    return ans;
}
std::vector<int> ans1, ans2;
int cnt1, cnt2;
inline void dfs(std::vector<int>& ans, int& cnt, int l, int r, int v) {
    if (l > r) {
        ++cnt;
        ans.push_back(v);
        return;
    }
    for (int i = 1; i <= m; ++i) {
        dfs(ans, cnt, l + 1, r, v + k[l] * mpow(i, p[l]));
    }
    return;
}
std::vector<int> cntans1;
int main() {
    scanf("%d%d", &n, &m);
    k.resize(n + 1);
    p.resize(n + 1);
    for (int i = 1; i <= n; ++i) {
        scanf("%d%d", &k[i], &p[i]);
    }
    dfs(ans1, cnt1, 1, n >> 1, 0);
    dfs(ans2, cnt2, (n >> 1) + 1, n, 0);
    std::sort(ans1.begin(), ans1.end());
    int newcnt1 = 1;
    cntans1.push_back(1);
    for (int i = 1; i < cnt1; ++i) {
        if (ans1[i] == ans1[newcnt1 - 1]) {
            ++cntans1[newcnt1 - 1];
        } else {
            ans1[newcnt1++] = ans1[i];
            cntans1.push_back(1);
        }
    }
    cnt1 = newcnt1;
    std::sort(ans2.begin(), ans2.end());
    int las = 0;
    ll ans = 0;
    for (int i = cnt2 - 1; i >= 0; --i) {
        for (; las < cnt1 && ans1[las] + ans2[i] < 0; ++las)
            ;
        if (las < cnt1 && ans1[las] + ans2[i] == 0)
            ans += cntans1[las];
    }
    printf("%lld\n", ans);
    return 0;
}

判断题

记程序结束前 $p$ 数组元素的最大值为 $P$，则该代码的时间复杂度是（ ）。

• A. $O(n)$
• B. $O(m^n \log m^n)$
• C. $O(m^{n/2} \log m^{n/2})$
• D. $O(m^{n/2} (\log m^{n/2} + \log P))$

{{ select(1) }}

• $O(n)$
• $O(m^n \log m^n)$
• $O(m^{n/2} \log m^{n/2})$
• $O(m^{n/2} (\log m^{n/2} + \log P))$