BestCoder Round #76 (div.2)

DZY Loves Partition

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1241    Accepted Submission(s): 463

思路：假设所给数是a，要分成n个数，最小是从1到n，sum=（1+n）*n/2,if(sum>a)输出-1；
否则一定是可解，假设先分成1~n；cnt（每个数的增量）=（a-sum）/n;j（余下的数）=（a-sum）%n，j分别给倒数j个数分别+1；应为题中说数不能重复。
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
24
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)

#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
#include <iostream>
#define LL long long
#include <climits>
using namespace std;
const LL MOD=1e9+7;
int main()
{
    int n;
    LL a,b;
    LL sum;
    cin>>n;
   while(n--)
   {
       cin>>a>>b;
       sum=(1+b)*b/2;
       if(sum>a)
       {
           printf("-1\n");
       }
       else
       {
           a-=sum;
           LL cnt=a/b;
           LL j=a%b;
           LL ans=1;
           for(LL i=b+cnt;i>=1+cnt;i--,j--)
           {
               if(j>0)
                (ans*=(i+1))%=MOD;
            else
                (ans*=(i))%=MOD;
           }
           printf("%I64d\n",ans);
       }

   }

    return 0;
}