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 partitionn 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 thesek 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

43 43 29 3666666 2

 

Sample Output

-1224110888111
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;}