HDU 5646

DZY Loves Partition

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 815    Accepted Submission(s): 298

Problem Description

DZY loves partitioning numbers. He wants to know whether it is possible to partition n

into the sum of exactly k

distinct positive integers.

After some thinking he finds this problem is Too Simple. So he decides to maximize the product of these k

numbers. Can you help him?

The answer may be large. Please output it modulo 109+7

.

 

Input

First line contains t

denoting the number of testcases.

t

testcases follow. Each testcase contains two positive integers n,k

in a line.

(1≤t≤50,2≤n,k≤109

)

 

Output

For each testcase, if such partition does not exist, please output −1

. Otherwise output the maximum product mudulo 109+7

.

 

Sample Input

4

3 4

3 2

9 3

666666 2

 

Sample Output

-1

2

2

4

110888111

Hint

In 1st testcase, there is no valid partition.
In 2nd testcase, the partition is $3=1+2$. Answer is $1\times 2 = 2$.
In 3rd testcase, the partition is $9=2+3+4$. Answer is $2\times 3 \times 4 = 24$. Note that $9=3+3+3$ is not a valid partition, because it has repetition.
In 4th testcase, the partition is $666666=333332+333334$. Answer is $333332\times 333334= 111110888888$. Remember to output it mudulo $10^9 + 7$, which is $110888111$.

 

Source

BestCoder Round #76 (div.2)

 

题意：t组数据 将n分为k个数的和 使得 这k个数的乘积最大（k个数不能重复） 输出这个数

官方题解

记sum(a,k)=a+(a+1)+⋯+(a+k−1)

首先，有解的充要条件是sum(1,k)≤n（如果没取到等号的话把最后一个k扩大就能得到合法解）。

然后观察最优解的性质，它一定是一段连续数字，或者两段连续数字中间只间隔1个数。这是因为1≤a<=b−2时有ab<(a+1)(b−1)如果没有满足上述条件的话，我们总可以把最左边那段的最右一个数字作为a，最右边那段的最左一个数字作为b，调整使得乘积更大。

可以发现这个条件能够唯一确定n的划分，只要用除法算出唯一的a使得sum(a,k)≤n<sum(a+1,k)就可以得到首项了。

理解我yan代码  先排列一个k位的首项为1 的等差数列 将剩下的数 优先靠右分配

       

 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cstdio>
 #include<map>
 using namespace std;
 #define mod 1000000007
 #define  ll __int64
 ll T,n,k,f;
 ll ans = ;
 int main() {
     scanf("%I64d",&T);
     while(T--)
     {
         scanf("%I64d%I64d",&n,&k);
         ans = ;
         if(k==)
         {
             cout<<n<<endl;
             continue;
         }
         if(n<k)
         cout<<-<<endl;
         else
         {
             n = n-(+k)*k/;//每个只能出现一次
             if(n<)
         {
                 cout<<-<<endl;continue;
         }
             if(n==)
             {
                 for(int i=;i<=k;i++)
                 ans=(ans*i)%mod;
                 cout<<ans<<endl;
                 continue;
             }
            for(int i=k;i>k-n%k;i--)
                ans=(ans*(n/k++i))%mod;
            for(int i=k-n%k;i>=;i--)
                 ans=(ans*(n/k+i))%mod;
            cout<<ans<<endl;
         }
     }
 }