HDU 5976 Detachment

Detachment

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2392    Accepted Submission(s): 682

Problem Description

In a highly developed alien society, the habitats are almost infinite dimensional space.
In the history of this planet,there is an old puzzle.
You have a line segment with x units’ length representing one dimension.The line segment can be split into a number of small line segments: a1,a2, … (x= a1+a2+…) assigned to different dimensions. And then, the multidimensional space has been established. Now there are two requirements for this space: 
1．Two different small line segments cannot be equal ( ai≠aj when i≠j).
2．Make this multidimensional space size s as large as possible (s= a1∗a2*...).Note that it allows to keep one dimension.That's to say, the number of ai can be only one.
Now can you solve this question and find the maximum size of the space?(For the final number is too large,your answer will be modulo 10^9+7)

 

Input

The first line is an integer T,meaning the number of test cases.
Then T lines follow. Each line contains one integer x.
1≤T≤10^6, 1≤x≤10^9

 

Output

Maximum s you can get modulo 10^9+7. Note that we wants to be greatest product before modulo 10^9+7.

 

Sample Input

14

 

Sample Output

4
题意：给你一个数，将这个数分成互不相等的若干个数是的这些数的乘积最大，求最大乘积。
对于一个数，最有情况是分成2,3,4,5...这种答案最优（不知道怎么证明）。这是就会有三种情况。
1：x=sum[n]，那么ans=n！。
2：x+1=sum[n]，这时去掉2，n加上1后得到的乘积最大。ans=n!/(2*n).
3:x+k=sum[n],k<n,k一定会抵消，ans=n!/(s[n]-x)。
计算式会用到乘法，需要使用逆元。
#include<stdio.h>  #include<algorithm>  #include<string.h>  using namespace std;#define ll long longconst int maxm = 100005;const int mod = 1e9 + 7;ll s[maxm], c[maxm];void init(){s[2] = 2, c[2] = 2;for (int i = 3;i <= 100000;i++){s[i] = s[i - 1] + i;c[i] = (c[i - 1] * i) % mod;}}int e_gcd(int a, int b, int &x, int &y){if (b == 0){x = 1;y = 0;return a;}int ans = e_gcd(b, a%b, x, y);int temp = x;x = y;y = temp - (a / b)*y;return ans;}int cal(int a, int m){int x, y, ans;int gcd = e_gcd(a, m, x, y);if (gcd != 1) return -1;m = abs(m);ans = x % m;if (ans <= 0) ans += m;return ans;}int main(){int ans, t, a, m, x, n;init();scanf("%d", &t);while (t--){scanf("%d", &x);if (x == 1){printf("1\n");continue;}n = lower_bound(s + 2, s + 2 + 100000, x) - s;//printf("%d %lld\n", n,s[n]);if (s[n] == x)printf("%lld\n", c[n]);else{if (s[n] - x == 1){ans = cal(2 * n, mod);printf("%lld\n", (c[n + 1] * ans)%mod);}else{ans = cal(s[n] - x, mod);printf("%lld\n", (c[n] * ans)%mod);}}}return 0;}