UVa 1363 POJ 2800 Joseph's Problem

Joseph likes taking part in programming contests. His favorite problem is, of course, Joseph’s problem.It is stated as follows.There are n persons numbered from 0 to n - 1 standing in a circle. The
person numberk, counting from the person number 0, is executed. After that the person number k of theremaining persons is executed, counting from the person after the last executed one. Theprocess 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 needis one cycle:r := 0;for i from 1 to n dor := (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 thefollowing algorithm:r := 0;for i from 1 to n dor := r + (k mod i);return r;Of course, Andrew pointed out that Joseph was wrong. But calculating the function Joseph
describedis also very interesting.Given n and k, find ∑ni=1(k mod i).

Input

The input file contains several test cases, each of them consists of a line containing n and k (1 ≤ n, k ≤109)

Output

For each test case, output a line containing the sum requested.

Sample Input

5 3

Sample Input

7

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（神奇的）找规律+数论~

做几次除法就能发现结果在一定区间内是递减的等差数列，如：

150%22=18，

150%23=12，

150%24=6，

150%25=0，

所以关键就成了求区间左右以及处理规律以外的数。

如果n大于k，多出的部分就是(n-k)*k；

除数从1到规律部分以外的部分直接加就可以了；

（其实说起来简单，算起来好麻烦的……）

（POJ上CE了，但在UVa上是A的……）

#include<cstdio>
#include<cmath>
#define ll long long

ll n,k;

ll cal()
{
ll a=(ll)sqrt(k),b=k/a,ans=0;
if(n>k) ans+=(n-k)*k;
for(ll i=a;i>1;i--)
{
ll e=k/i,f=k/(i-1);
if(e>n) break;
if(f>n) f=n;
ans+=((k%f)+(k%(e+1)))*(f-e)/2;
}
for(ll i=1;i<=n && i<=b;i++) ans+=k%i;
return ans;
}

int main()
{
while(scanf("%lld%lld",&n,&k)!=EOF) printf("%lld\n",cal());
return 0;
}