A. Slackline Adventure

考虑枚举每个矩形的长跟宽，再统计这个矩形能在坐标轴上出现几次（同行相邻同列相邻的单独算），然后有如下式子：

2 × ∑ i = 1 n − 1 ∑ j = 1 m − 1 ( n − i ) ( m − j ) [ gcd ⁡ ( i , j ) = 1 ] [ L 2 ≤ i 2 + j 2 ≤ R 2 ] 2 \times \sum_{i = 1} ^{n - 1} \sum_{j = 1} ^{m - 1}(n - i)(m - j)[\gcd(i, j) = 1][L ^ 2 \leq i ^ 2 + j ^ 2 \leq R ^ 2]\\ 2×i=1∑n−1​j=1∑m−1​(n−i)(m−j)[gcd(i,j)=1][L2≤i2+j2≤R2]
设 F ( n , m , T 2 ) = ∑ i = 1 n − 1 ∑ j = 1 m − 1 ( n − i ) ( m − j ) [ gcd ⁡ ( i , j ) = 1 , i 2 + j 2 ≤ T 2 ] F(n, m, T ^ 2) = \sum\limits\limits_{i = 1} ^{n - 1} \sum\limits_{j = 1} ^{m - 1} (n - i)(m - j)[\gcd(i, j) = 1, i ^ 2 + j ^ 2 \leq T ^ 2] F(n,m,T2)=i=1∑n−1​j=1∑m−1​(n−i)(m−j)[gcd(i,j)=1,i2+j2≤T2]，则 a n s = F ( n , m , R 2 ) − F ( n , m , L 2 − 1 ) ans = F(n,m, R ^ 2) - F(n, m, L ^ 2 - 1) ans=F(n,m,R2)−F(n,m,L2−1)。

∑ k = 1 n − 1 μ ( k ) ∑ i = 1 n − 1 k ∑ j = 1 m − 1 k ( n − i k ) ( m − j k ) [ i 2 k 2 + j 2 k 2 ≤ T 2 ] ∑ k = 1 n − 1 μ ( k ) ∑ i = 1 n − 1 k ( n − i k ) c a l c ( ) \sum_{k = 1} ^{n - 1} \mu(k) \sum_{i = 1} ^{\frac{n - 1}{k}} \sum_{j = 1} ^{\frac{m - 1}{k}} (n - ik)(m - jk)[i ^ 2 k ^ 2 + j ^ 2 k ^ 2 \leq T ^ 2]\\ \sum_{k = 1} ^{n - 1} \mu(k) \sum_{i = 1} ^{\frac{n - 1}{k}} (n - ik) calc()\\ k=1∑n−1​μ(k)i=1∑kn−1​​j=1∑km−1​​(n−ik)(m−jk)[i2k2+j2k2≤T2]k=1∑n−1​μ(k)i=1∑kn−1​​(n−ik)calc()
c a l c ( ) calc() calc()就是当已知 i , k , T i, k, T i,k,T，有多少 j j j满足 i 2 k 2 + j 2 k 2 ≤ T 2 i ^ 2 k ^ 2 + j ^ 2 k ^ 2 \leq T ^ 2 i2k2+j2k2≤T2，的 m − j k m - jk m−jk求和，整体复杂度 O ( k × n log ⁡ n ) O(k \times n \log n) O(k×nlogn)， k k k是个比较小的常数

今天VP的时候碰到的，到了一直没看这题，到了最后才发现是个数论题。

说一点点坑点吧，注意函数不能写成 f ( n , m , T ) = ∑ i = 1 n − 1 ∑ j = 1 m − 1 ( n − i ) ( m − j ) [ gcd ⁡ ( i , j ) = 1 , i 2 + j 2 ≤ T 2 ] f(n, m, T) = \sum\limits\limits_{i = 1} ^{n - 1} \sum\limits_{j = 1} ^{m - 1} (n - i)(m - j)[\gcd(i, j) = 1, i ^ 2 + j ^ 2 \leq T ^ 2] f(n,m,T)=i=1∑n−1​j=1∑m−1​(n−i)(m−j)[gcd(i,j)=1,i2+j2≤T2]，

然后答案是 f ( n , m , R ) − f ( n , m , L − 1 ) f(n, m, R) - f(n, m, L - 1) f(n,m,R)−f(n,m,L−1)，这样会使满足条件得满足条件的一些数 ( L − 1 ) 2 < i 2 k 2 + j 2 k 2 < L 2 (L - 1) ^ 2 <i ^ 2 k ^ 2 + j ^ 2 k ^ 2 < L ^ 2 (L−1)2<i2k2+j2k2<L2，

没有被删去而算在贡献里了，会导致答案出错，就因为这个调了一晚上，一直不知道哪里错了，，，

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 10, mod = 1e9 + 7;

int prime[N], mu[N], cnt, n, m, L, R;

bool st[N];

inline int add(int x, int y) {
  return x + y < mod ? x + y : x + y - mod;
}

inline int sub(int x, int y) {
  return x >= y ? x - y : x - y + mod;
}

void init() {
  mu[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!st[i]) {
      mu[i] = -1;
      prime[++cnt] = i;
    }
    for (int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {
      st[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
  }
}

int calc(int i, int k, long long T) { // i * i * k * k + j * j * k * k <= T
  int l = 0, r = (m - 1) / k;
  while (l < r) {
    int mid = l + r + 1 >> 1;
    if (1ll * i * i * k * k + 1ll * mid * mid * k * k > T) {
      r = mid - 1;
    }
    else {
      l = mid;
    }
  }
  return 1ll * (m - k + m - l * k) * l / 2 % mod;
}

int f(long long T) {
  int ans = 0;
  for (int k = 1, lim, res; k <= n - 1; k++) {
    if (mu[k] == 0) {
      continue;
    }
    lim = (n - 1) / k, res = 0;
    for (int i = 1; i <= lim; i++) {
      res = add(res, 1ll * (n - i * k) * calc(i, k, T) % mod);
    }
    if (mu[k] == 1) {
      ans = add(ans, res);
    }
    else {
      ans = sub(ans, res);
    }
  }
  ans = 2ll * ans % mod;
  if (T >= 1) {
    ans = add(ans, add(1ll * n * (m - 1) % mod, 1ll * (n - 1) * m % mod));
  }
  return ans;
}

int main() {
  // freopen("in.txt", "r", stdin);
  // freopen("out.txt", "w", stdout);
  init();
  scanf("%d %d %d %d", &n, &m, &L, &R);
  printf("%d\n", sub(f(1ll * R * R), f(1ll * L * L - 1)));
  return 0;
}