hdu5646 DZY Loves Partition 数学

ZY Loves Partition

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 365    Accepted Submission(s): 138

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.

 

题意：给定数n和数k，使k个不同的数的和味n，且要求这k个数的乘积尽可能大，问乘积对1e9+7取模后是多少。

解:若sum(1,k)已经大于n则必无解，否则可通过对最大数增加一定能得到解。

当有解时：由基本不等式的(a1+a2+...+an)*1/n>=(a1*a2...*an)^1/n.挡a1,a2...相等时取等号，要使乘积尽可能大，那么需要这n个数尽可能接近，于是求连续最大和sum(begin,begin+k-1)<=n，然后把n-sum(begin.begin+k-1)剩下的从大到小分配即可。

AC代码：

//************************************************************************////*Author : Handsome How                                                 *////************************************************************************////#pragma comment(linker, "/STA    CK:1024000000,1024000000")#pragma warning(disable:4996) #include <vector>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <algorithm>#include <sstream>#include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <ctime>                                                #include <cassert>#if defined(_MSC_VER) || __cplusplus > 199711L#define aut(r,v) auto r = (v)#else#define aut(r,v) __typeof(v) r = (v)#endif#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)#define fur(i,a,b) for(int i=(a);i<=(b);i++)#define furr(i,a,b) for(int i=(a);i>=(b);i--)#define cl(a) memset((a),0,sizeof(a))using namespace std;typedef long long LL;//----------------------------------------------------const LL mod = 1e9 + 7;int main(){    //freopen("E:\\data.in", "r", stdin);    ios :: sync_with_stdio(false);    int T;    scanf("%d", &T);    while (T--)    {        LL n, k;        scanf("%I64d%I64d", &n, &k);        LL t;        t = (1 + k)*k / 2;        if(t>n) {            printf("-1\n");            continue;        }        LL begin;        begin = (2 * n / k + 1 - k) / 2;        LL res =n - (begin + begin + k - 1)*k / 2;        LL real = 1;        LL end = begin + k - 1;        while (res)        {            real *= (end + 1);            real %= mod;            res--;            end--;        }        for (LL i = begin; i <= end; i++) {            real *= i;            real %= mod;        }        printf("%I64d\n",real);    }    return 0;}