HDU-2620 （数学，循环节）

最新推荐文章于 2021-02-15 23:41:08 发布

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最新推荐文章于 2021-02-15 23:41:08 发布

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本文链接：https://blog.youkuaiyun.com/l18339702017/article/details/80011833

本文介绍了一种解决特定数论问题的算法实现，通过计算给定整数n和k的函数Ice(n,k)，该函数涉及到求和运算及等差数列的分段累加。代码使用C++编写，展示了如何高效地解决这一问题。

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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.

InputGiven two integers n, k(1 <= n, k <= 10 ⁹).OutputFor each n and k, print Ice(n, k) in a single line.Sample Input

5 4
5 3

Sample Output

5
7

题意：求和k%i (1<=i<=n)

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。

#include<bits/stdc++.h>
#define LL long long
using namespace std;

int main(){
    LL n,k;
    while(scanf("%lld%lld",&n,&k)!=EOF){
        LL ans = n*k;
        if(n > k)   n = k;
        for(int i=1;i<=n;){
            LL d = k/i;
            LL j = k/d;
            if(j > n) j = n;
            LL temp = (i+j)*(j-i+1)/2*d;
            ans -= temp;
            i = j+1;
        }
        cout<<ans<<endl;
    }
    return 0;
}