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
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int N = 1e5 + 5; 5 const int mod = 1e9 + 7; 6 ll mul[N]; //前缀积 7 ll sum[N]; //前缀和 8 void init() { //初始化 9 mul[1] = 1; 10 sum[1] = 0; 11 for (int i = 2; i < N; i++) { 12 sum[i] = sum[i - 1] + i; //前缀和 13 mul[i] = (i * mul[i - 1]) % mod; //前缀积 14 } 15 } 16 ll inv(ll a, ll b) { 17 ll ans = 1; 18 while (b) { 19 if (b & 1) { 20 ans = (ans * a) % mod; 21 } 22 a = (a * a) % mod; 23 b >>= 1; 24 } 25 return ans; 26 } 27 int main() { 28 ios::sync_with_stdio(false); 29 cin.tie(0); 30 cout.tie(0); 31 init(); 32 int t; 33 cin >> t; 34 while (t--) { 35 int x; 36 cin >> x; 37 if (x == 1) { 38 cout << 1 << endl; 39 continue; 40 } 41 int l = 1, r = N + 1; 42 while (l < r) { 43 int mid = (l + r + 1) / 2; 44 if (sum[mid] <= x) { 45 l = mid; 46 } else { 47 r = mid - 1; 48 } 49 } 50 //cout << l << endl; 51 int p = l; 52 int num = x - sum[p]; //num就是相差的值 53 ll ans = 0; 54 if (num == l) { //如果num刚好和k相等 55 // 2 3 4 5 6 7 7 56 // 3 4 5 6 7 9 57 ans = (mul[p] * inv(2, mod - 2) % mod * (p + 2)) % mod; 58 } else { 59 ans = (mul[p + 1] * inv(mul[p + 1 - num], mod - 2) % mod * mul[p - num]) % mod; 60 } 61 cout << ans << endl; 62 } 63 return 0; 64 }