2016年ICPC亚洲区域赛大连站F题Detachment

题目链接https://ac.nowcoder.com/acm/problem/124645

题意简介：

　　把给定的数x，拆成若干个不相等的数相加，x = a1 + a2 + ...，使得这些不相等的数的乘积s = a1 * a2 * ...最大，求最大的s

解题思路：

　　参考自https://www.cnblogs.com/ymzjj/p/9865052.html

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
ll mul[N]; //前缀积 
ll sum[N]; //前缀和 
void init() { //初始化 
    mul[1] = 1;
    sum[1] = 0;
    for (int i = 2; i < N; i++) {
        sum[i] = sum[i - 1] + i; //前缀和
        mul[i] = (i * mul[i - 1]) % mod; //前缀积
    }
}
ll inv(ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) {
            ans = (ans * a) % mod;
        }
        a = (a * a) % mod;
        b >>= 1;
    }
    return ans;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    init();
    int t;
    cin >> t;
    while (t--) {
        int x;
        cin >> x;
        if (x == 1) {
            cout << 1 << endl;
            continue;
        }
        int l = 1, r = N + 1; 
        while (l < r) {
            int mid = (l + r + 1) / 2;
            if (sum[mid] <= x) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        //cout << l << endl;
        int p = l;
        int num = x - sum[p]; //num就是相差的值 
        ll ans = 0;
        if (num == l) { //如果num刚好和k相等 
            // 2 3 4 5 6 7 7
            // 3 4 5 6 7 9
            ans = (mul[p] * inv(2, mod - 2) % mod * (p + 2)) % mod;
        } else {
            ans = (mul[p + 1] * inv(mul[p + 1 - num], mod - 2) % mod * mul[p - num]) % mod;
        }
        cout << ans << endl;
    }
    return 0;
}