Codeforces Round #701 (Div. 2) C. Floor and Mod 整除分块

题意

给一个数对(n, m)，求出在a∈[1,n]，b∈[1,m]中满足$\lfloor$$\frac{a}{b}$$\rfloor$ = a mod b的数对有多少个

分析

观察样例发现当a=b+1时肯定满足，然后接着往下推

当b=2时
a / b = 3 / 2 … 再往后$\lfloor$$\frac{a}{b}$$\rfloor$ 已经大于2了，因此不可能满足

当b=3时
a / b = 4 / 3, 8 / 3

当b=4时
a / b = 5 / 4, 10 / 4, 15 / 4
…
很容易发现当b = m时，最多只有m-1个a与之对应

然后观察a，发现每个a之间是相差了b+1，计算个数就可以写成$\lfloor$$\frac{n}{b+1}$$\rfloor$

所以可以考虑去枚举b，但一个个去枚举肯定超时，所以想到整除分块

写成式子就是${\sum }_{i=2}^{b}min(i-1,n/(i+1))$其中b为min(n-1, m)

Code

#include <bits/stdc++.h>
using namespace std;
//#define ACM_LOCAL
#define fi first
#define se second
#define il inline
#define re register
const int N = 5e5 + 10;
const int M = 5e5 + 10;
const int INF = 0x3f3f3f3f;
const double eps = 1e-5;
const int MOD = 10007;
typedef long long ll;
typedef pair<int, int> PII;
typedef unsigned long long ull;

ll calc(ll n, ll m) {
    
    
    ll ans = 0;
    for (ll l = 2, r; l <= min(n-1, m); l = r + 1) {
    
    
        r = min(min(n-1, m), n / (n / (l+1)) - 1);//注意这里的-1
        ans = ans + min(r*(r-1)/2 - (l-1)*(l-2)/2, (r - l + 1) * (n / (l+1)));
    }
    return ans;
}

void solve() {
    
    
    int T; cin >> T; while (T--) {
    
    
        ll n, m; cin >> n >> m;
        if (n <= 2 || m < 2) cout << 0 << endl;
        else {
    
    
            cout << calc(n, m) << endl;
        }
    }
}

signed main() {
    
    
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
#endif
    solve();
}