HDU 5976 Detachment（拆分）
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Problem Description - 题目描述
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)
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)
在一个高度发达的外星文明中，有着近乎无限维度的生存空间。
在这颗星球的历史中，有道古老的谜题。
你有一条长x个单位长度的线段表示一个维度。这条线段可以被拆分为若干小线段：a1,a2, … (x= a1+a2+…)并分配为不同的维度。然后，多维空间就建立起来了。现在，这个空间有两个限制：
．两个不同的小线段不能相等( ai≠aj when i≠j)。
．多维空间的大小s要尽可能大(s= a1∗a2*...)。注意，各维度只能保持一种。也就是说，ai的值必须唯一。
现在的你能解决这个问题并找出最大的空间吗？（结果可能很大，输出模10^+）
CN
Input - 输入
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
Then T lines follow. Each line contains one integer x.
1≤T≤10^6, 1≤x≤10^9
第一行为一个整数T，描述测试用例的数量。
随后T行。每行有一个整数x。
≤T≤^, ≤x≤^
CN
Output - 输出
Maximum s you can get modulo 10^9+7. Note that we wants to be greatest product before modulo 10^9+7.
s的最大值需要模10^+。注意，模10^+7是在获得最大乘积后。
CN
Sample Input - 输入样例
1
4
Sample Output - 输出样例
4
题解
先猜一发最优策略：2+3+4+5+6+……
然后再猜一发对于剩下数的分配策略：每次从后往前，对每个数+1。
接着就发现似乎策略就是这样了。
后面需要做的处理：求前n项和，求前n项积。
最后遇到（a % mod）/（b % mod）的时候需要用逆元。
把（a/b）%mod转化为（a * inv b）%mod 不嫌弃速度的话可以用费马小定理：
mod为质数时，inv a = a^(mod - 2)
或者用其他方法…………
代码 C++
#include <cstdio>
#include <algorithm>
#define mod 1000000007
#define mx 44722
__int64 mul[mx] = { }, sum[mx];
__int64 qMod(__int64 a, int n){
__int64 opt = ;
while (n){
if (n & ) opt = (opt*a) % mod;
n >>= ;
a = (a*a) % mod;
}
return opt;
}
__int64 lr(int l, int r){//[l, r]
return (mul[r] * qMod(mul[l - ], mod - )) % mod;
}
void rdy(){
int i, j;
for (i = , j = ; i < mx; ++i, ++j){
sum[i] = j + sum[i - ];
mul[i] = (j * mul[i - ]) % mod;
}
}
int main(){
rdy();
int t, len, w, l, r;
__int64 x, opt;
scanf("%d", &t);
while (t--){
scanf("%I64d", &x);
if (x < ) opt = x;
else{
len = std::upper_bound(sum, sum + mx, x) - sum - ;
r = len + (x - sum[len]) / len;
w = (x - sum[len]) % len;
opt = lr(r - len + , r - w);
if (w) opt *= lr(r + - w, r + ), opt %= mod;
}
printf("%I64d\n", opt);
}
return ;
}