Ultraviolet A 12508-Triangles in the grid (ry + count)

Link: Ultraviolet A 12508-Triangles in the grid

N, m, A, and B are given, which must be calculated in (n + 1 )? On the (m + 1) matrix, you can find the number of triangles with an area between AB.

Solution: first, enumerate the matrix, and then calculate the number of triangles using this matrix as an external matrix, and the size must be between AB. Then, for each matrix, you can determine a few within a large range.

The inner triangle of the enumeration matrix can be divided into three types:
1. The two points of the triangle shape are located at the vertex of a matrix edge, and the other point is located at the peer edge of the edge (excluding the vertex)

2. Side of a triangle with diagonal lines

In this way, we can enumerate X and then obtain the boundary values of L and R.

3. The triangle points are on the vertex of the rectangle, and the other two points are on the corresponding edge.

X is also enumerated, but this time x cannot include 0 and N (calculated in Case 2), corresponding to the red triangle and the blue triangle, the area is reduced by X, therefore, we can calculate the number of triangles.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;inline ll max(ll a, ll b) {    return a > b ? a : b;}inline ll min(ll a, ll b) {    return a < b ? a : b;}ll N, M, A, B;ll solve (ll k) {    if (k < 0)        k = 0;    if (N > M)        swap(N, M);    ll ans = 0;    for (ll n = 1; n <= N; n++) {        for (ll m = 1; m <= M; m++) {            ll cnt = 0;            if (n * m <= k)                cnt += 2 * (n + m - 2);            ll l, r;            for (ll x = 0; x <= n; x ++) {                r = (m * x + k) / n;                if (r > m)                    r = m;                ll t = m * x - k;                if(t <= 0)                     l = 0;                else                     l = (t - 1) / n + 1;                if(l <= r)                     cnt += 2 * (r - l + 1);            }            for (ll x = 1; x < n; x++) {                ll tmp = n * m - x;                if (tmp <= k)                    cnt += 4 * (m - 1);                else {                    tmp = tmp - k;                    ll u = m-1 - min(tmp / x + (tmp % x != 0), m-1);                    cnt += 4 * u;                }            }            ans += cnt * (N - n + 1) * (M - m + 1);        }    }    return ans;}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        scanf("%lld%lld%lld%lld", &N, &M, &A, &B);        printf("%lld\n", solve(B*2) - solve(A*2-1));    }    return 0;}