2015 ACM ICPC Singapore Regional D（折半枚举+二分）

D - Association of Computer Maintenance

题意

给定至多350个小于100的质数，对于所有质数之积k将它分解为两个数a和b使得a*b=k。输出最小的a+b，并对1e9+7取模

分析

首先考虑想如果想让a+b最小，即让abs(a-b)最小。根据限制条件k的因子数不超过1e10，容易想到将k拆分成k1和k2，此时对于k1和k2的因子数都不超过1e5。考虑将k1的所有因子存下，最后再枚举k2的所有因子去匹配最接近的k1因子，记录下最小值即可。
注意维护因子时由于高精度过慢，考虑转换成对数比较大小。时间复杂O(nlogn)。

代码实现

#include <bits/stdc++.h>
int main() {
    std::cin.tie(nullptr)->sync_with_stdio(false);
    #define int long long
    constexpr int P = 1e9 + 7;
    std::string str;
    std::cin >> str;
    std::vector<int> pri;
    for (int i = 0; i < (int)size(str); i += 2) {
        pri.emplace_back((str[i] - '0') * 10 + (str[i + 1] - '0'));
    }
    std::sort(pri.begin(), pri.end());
    std::vector<std::pair<int, int>> fac;
    for (int i = 0; i < (int)size(pri); ++i) {
        int j = i + 1;
        while (j < (int)size(pri) && pri[i] == pri[j]) j += 1;
        fac.emplace_back(pri[i], j - i);
        i = j - 1;
    }
    int pos = (int)size(fac) - 1;
    for (int i = 0, j = 1; i < (int)size(fac); ++i) {
        if (j * (fac[i].second + 1) > 4e5) {
            pos = i - 1;
            break;
        } else {
            j *= fac[i].second + 1;
        }
    }
    auto power = [&](int a, int b) {
        int c = 1;
        for (; b; b >>= 1, a = a * a % P)
            if (b & 1) c = c * a % P;
        return c;
    };
    std::vector<std::pair<int, int>> s;
    std::vector<std::pair<double, int>> prod;
    int idx = 0;
    auto dfs = [&](auto &&self, int u, double diff, int s1, int s2) ->void {
        if (u == pos + 1) {
            prod.emplace_back(diff, idx++);
            s.emplace_back(s1, s2);
            return ;
        }
        auto [p, cnt] = fac[u];
        for (int i = 0; i <= cnt; ++i) {
            self(self, u + 1, diff + log(p) * (i + i - cnt), s1 * power(p, i) % P, s2 * power(p, cnt - i) % P);
        }
    };
    dfs(dfs, 0, 0, 1, 1);
    std::sort(prod.begin(), prod.end());
    int ans = 0;
    double max = 1e100;
    auto find = [&](auto &&self, int u, double diff, int s1, int s2) ->void {
        if (u == (int)size(fac)) {
            int p = std::lower_bound(prod.begin(), prod.end(), std::pair<double, int>{-diff, -1}) - prod.begin();
            if (p < (int)size(prod)) {
                auto [w, id] = prod[p];
                if (fabs(diff + w) < max) {
                    max = fabs(diff + w);
                    ans = ( s1 * s[id].first % P +  s2 * s[id].second % P) % P;
                }
            }
            if (p - 1 >= 0) {
                auto [w, id] = prod[p - 1];
                if (fabs(diff + w) < max) {
                    max = fabs(diff + w);
                    ans = ( s1 * s[id].first % P +  s2 * s[id].second % P) % P;
                }
            }
            return ;
        }
        auto [p, cnt] = fac[u];
        for (int i = 0; i <= cnt; ++i) {
            self(self, u + 1, diff + log(p) * (i + i - cnt), s1 * power(p, i) % P, s2 * power(p, cnt - i) % P);
        }
    };
    find(find, pos + 1, 0, 1, 1);
    std::cout << ans << '\n';
}