HDU 5646 DZY Loves Partition（二分法）

DZY Loves Partition

 

 Accepts: 154

 

 Submissions: 843

 Time Limit: 4000/2000 MS (Java/Others)

 

 Memory Limit: 262144/262144 K (Java/Others)

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 10^9+710​9​​+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\le t\le 50, 2\le n,k \le 10^91≤t≤50,2≤n,k≤10​9​​)

Output

For each testcase, if such partition does not exist, please output $-1$. Otherwise output the maximum product mudulo 10^9 + 710​9​​+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 + 710​9​​+7, which is 
      
      
      
      $110888111$.

     
     
     
     
     
     
     
     


     
     
     
     
     
     
     
     


     
     
     
     
     
     
     
     

      
      
      
      大体题意：
     
     
     
     
     
     
     
     

      
      
      
      给你个数n，拆k个各不相同的数之和，并且使得k个数的乘积最大，最后输出k个数乘积对1e9+7取模的结果！
     
     
     
     
     
     
     
     


     
     
     
     
     
     
     
     

      
      
      
      思路：
     
     
     
     
     
     
     
     

      
      
      
      令sum(a,b),为a + （a + 1） + （a + 2）....+b的结果！
     
     
     
     
     
     
     
     

      
      
      
      先取得的一个数a使得sum(a,a+k-1)刚好小于n，
     
     
     
     
     
     
     
     

      
      
      
      最后多出来的数字加到这一串数后面，每个数加1就行了！
     
     
     
     
     
     
     
     

      
      
      
      通过二分来查找最大的a
     
     
     
     
     
     
     
     

      
      
      
      二分：
     
     
     
     
     
     
     
     

      
      
      
      先对n取半，如果此时sum(a,a+k-1)比n大，则向左找，否则向右找，并且记录最大值MAX
     
     
     
     
     
     
     
     

      
      
      
      如果最后发现MAX  没变的话，则不存在一个a满足式子，则直接输出-1
     
     
     
     
     
     
     
     

      
      
      
      否则就用那个a即可！
     
     
     
     
     
     
     
     


     
     
     
     
     
     
     
     

      
      
      
      
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
ll sum(ll a,ll b){
    ll num = b-a+1;
    ll ans;
    ans = num%2 ? (a+b)/2*num : num/2*(a+b);
    return ans;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        ll n,k;
        scanf("%lld%lld",&n,&k);
        ll MAX = -1;
        ll a,l=1,r=n;
        while(l <= r){
            ll mid =(r+l)/2;
            ll tmp = sum(mid,mid+k-1);
            if (tmp > n)r = mid-1;
            if (tmp <= n){
                l = mid + 1;
                if (tmp > MAX){
                    MAX = tmp;
                    a = mid;
                }
            }
        }
        if (MAX == -1){
            printf("-1\n");
            continue;
        }
        ll tp = n-MAX;
        ll ans = 1;
        for (int i = a; i < a+k; ++i){
            if (i >= a+k-tp)ans=(ans%mod*((i+1)%mod)%mod)%mod;
            else ans = (ans %mod * i%mod)%mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
}