codeforces 1157 D. N Problems During K Days (二分+构造)

D. N Problems During K Days
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Polycarp has to solve exactly n problems to improve his programming skill before an important programming competition. But this competition will be held very soon, most precisely, it will start in k days. It means that Polycarp has exactly k days for training!

Polycarp doesn’t want to procrastinate, so he wants to solve at least one problem during each of k days. He also doesn’t want to overwork, so if he solves x problems during some day, he should solve no more than 2x problems during the next day. And, at last, he wants to improve his skill, so if he solves x problems during some day, he should solve at least x+1 problem during the next day.

More formally: let [a1,a2,…,ak] be the array of numbers of problems solved by Polycarp. The i-th element of this array is the number of problems Polycarp solves during the i-th day of his training. Then the following conditions must be satisfied:

sum of all ai for i from 1 to k should be n;
ai should be greater than zero for each i from 1 to k;
the condition ai<ai+1≤2ai should be satisfied for each i from 1 to k−1.
Your problem is to find any array a of length k satisfying the conditions above or say that it is impossible to do it.

Input
The first line of the input contains two integers n and k (1≤n≤109,1≤k≤105) — the number of problems Polycarp wants to solve and the number of days Polycarp wants to train.

Output
If it is impossible to find any array a of length k satisfying Polycarp’s rules of training, print “NO” in the first line.

Otherwise print “YES” in the first line, then print k integers a1,a2,…,ak in the second line, where ai should be the number of problems Polycarp should solve during the i-th day. If there are multiple answers, you can print any.

昨天div3 的D题，题目很简单，题意：给你一个N，和长度为K的数组，数组要符合a[i] < a[i+1] < 2*a[i],请你输出任意一个数组和为N的数组。
这个题要注意起点即a[1]可以不是1。
思路：首先想到序列最小和肯定是a[1],a[1]+1,a[1]+2,…等差数列，最大和肯定是等比数列，如果知道他的起点的数，我们就能得到最小最大的和。所以我们可以二分一下起点就可以了。然后从后面往前补就行了，就是当前这个数能够到达的最大的数是前一个数的二倍。
注意这里有个坑点，如果K大于一定的数，那么起点一定是1，我这里是K>32的时候。
代码如下：

#include<bits/stdc++.h>

using namespace std;
const int MAX = 1e5+10;
typedef long long ll;
ll a[MAX];
ll N,K;
ll sumMin(