UVA1363

题意：给出n和k，1≤n,k≤1e9，计算

切入点是k/i 和 k/(i+1)差距不大。令p_i = k/i, r_i = k%i。如果p_i+1 == pi，那么r_i+1 == k - p_i(i+1) == r_i - p_i，

对于p_i+z == p_i，r_i+z == r_i - z*p_i，这是等差数列可以O(1)计算出来。如果上界为j，那么k/j ≤ p_i，j ≤ k/p_i。

/*********************************************************
*      --------------Tyrannosaurus---------              *
*   author AbyssalFish                                   *
**********************************************************/
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;


ll solve(int n, int k)
{
    ll ans = 0;
    for(int i = 1; i <= n; ){
        int p = k/i;
        if(!p){
            ans += (n+1-i)*(ll)k;
            break;
        }
        int j = min(k/p, n), r = k-i*p;
        ans += (2LL*r - (j-i)*p)*(j-i+1LL)/2LL;
        i = j+1;
    }
    return ans;
}

//#define LOCAL
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    int n, k;
    while(~scanf("%d%d",&n,&k)){
        printf("%lld
", solve(n,k));
    }
    return 0;
}