【HDU - 2620】Ice Rain （数学+分块+取余）

Ice Rain------I was waiting for a girl, or waiting for been addicted to the bitter sea. Love for irrigation in silence. No one considered whether the flowers came out or wither. Love which I am not sure swing left and right. I had no choice but to put my sadness into my heart deeply. 

   Yifenfei was waiting for a girl come out, but never. 
His love is caught by Lemon Demon. So yifenfei ’s heart is “Da Xue Fen Fei” like his name. 
The weather is cold. Ice as quickly as rain dropped. Lemon said to yifenfei, if he can solve his problem which is to calculate the value of  , he will release his love. 
Unluckily, yifenfei was bored with Number Theory problem, now you, with intelligent, please help him to find yifenfei’s First Love. 

Input

Given two integers n, k(1 <= n, k <= 10 9).

Output

For each n and k, print Ice(n, k) in a single line.

Sample Input

5 4
5 3

Sample Output

5
7

题意：

给n和k，问k对所有小于等于n并且大于等于1的数取余后的值得和是多少。

思路：

对这道题，想到了大于k的部分直接乘以个数，小于的部分本来想找规律，最后看了别人的题解发现是通过推公式的。。。

以下推理转自该博客:

对于每一个i     ,k%i = k- (k/i)*i；

则有： sum = [k-(k/1)*1]+[k-(k/2)*2]+······[k-(k/n)*n] = n*k  -   [(k/1)*1+(k/2)*2+······+(k/n)*n]；

n*K 可直接求得 

对于  [(k/1)*1+(k/2)*2+······+(k/n)*n]，因为 k  / i   是分段的，且，每一段都是规则的等差数列，我们可以分段累加。

ac代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<string>
#include<vector>
using namespace std;
typedef long long ll;

int main()
{
	ll n,k;
	while(~scanf("%lld%lld",&n,&k))
	{
		ll ans=n*k;
		if(n>k) n=k;//比k大的部分经过除法都变成了0
		for(int i=1;i<=n;)
		{
			ll d=k/i;//求出k/i的值，也就是这一段的值 
			ll j=k/d;//求这一段的终点？？
			if(j>n) j=n;
			ll tmp=(i+j)*(j-i+1)/2*d;//k/i是一样的，提出来，后面是求一段区间的和，用求和公式，然后乘k/i
			ans-=tmp;
			i=j+1;//跳到下一段 
		} 
		printf("%lld\n",ans); 
	}
	
}