给n 和 k 求: ∑1<=i<=n(k
mod i).
p = k/i
k mod i = k - p * i
k mod ( i + 1 ) = k - p * ( i + 1 ) = k mod i - p
k mod ( i + 2 ) = k - p * ( i + 2 ) = k mod i - 2 * p
对于连续的 i ,很多p都是一样的 . 相差的部分是一个等差数列 ,
i 的 范围是 从 i 到 min(k/p,n) 如果 p == 0 则 一直延续到最后
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Joseph‘s Problem
Description Joseph likes taking part in programming contests. His favorite problem is, of course, Joseph‘s problem. It is stated as follows.
The process continues until only one person is left. This person is a survivor. The problem is, given n and k detect the survivor‘s number in the original circle. Of course, all of you know the way to solve this problem. The solution is very short, all you need is one cycle: r := 0; for i from 1 to n do r := (r + k) mod i; return r; Here "x mod y" is the remainder of the division of x by y, But Joseph is not very smart. He learned the algorithm, but did not learn the reasoning behind it. Thus he has forgotten the details of the algorithm and remembers the solution just approximately. He told his friend Andrew about the problem, but claimed that the solution can be found using the following algorithm: r := 0; for i from 1 to n do r := r + (k mod i); return r; Of course, Andrew pointed out that Joseph was wrong. But calculating the function Joseph described is also very interesting. Given n and k, find ∑1<=i<=n(k mod i). Input The input file contains n and k (1<= n, k <= 109). Output Output the sum requested. Sample Input 5 3 Sample Output 7 Source |
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/* ***********************************************
Author :CKboss
Created Time :2015年02月02日 星期一 10时53分55秒
File Name :POJ2800.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long int LL;
LL n,k;
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(cin>>n>>k)
{
LL ret=0;
for(int i=1;i<=n;i++)
{
LL p = k/i;
if(p==0)
{
ret += (n-i+1)*k;
break;
}
else
{
LL q = k/p;
if(q>n) q = n;
LL md = k%i;
LL num = q-i+1;
LL temp = md*num-(num-1)*num/2*p;
ret += temp;
i=q;
}
}
cout<<ret<<endl;
}
return 0;
}
POJ 2800 Joseph's Problem