hdu 5646 DZY Loves Partition(整数拆分)

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

题意：

给定n,问能否把n拆成恰好k个不重复的正整数之和。可以的话求这k个正整数的乘积的最大值

solution：

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  1≤a<=b−2时 有ab<(a+1)(b-1)ab<(a+1)(b−1)

用sum代表1~k的和，有解的情况是sum<=n,现在看是连续数字还是两段数字间差一个数,设u=(n-sum)/k,v=(n-sum)%k.如果v==0的话，那么答案就为1~k每个数都加上u，若v！=0,那么1~k前k-v加u,剩下的加u+1即可。

#include<cstdio>#include<iostream>using namespace std;typedef long long ll;const int mod = 1e9 + 7;ll n, k;int main(){int t;;scanf("%d", &t);while (t--){cin >> n >> k;ll sum = k*(k + 1) / 2;if (sum > n)printf("-1\n");else {ll u = (n-sum)/ k, v = (n-sum)%k;long long ans = 1;for (ll i = 1; i <= k; i++)if (i <= k - v)ans = (ans*(i + u)) % mod;else ans = (ans*(i + u + 1)) % mod;cout << ans << endl;}}return 0;}