题目大意:给你$a,b,c,d,k(1leqslant a,b,c,d,kleqslant 5 imes10^4)$,求$displaystylesumlimits_{x=a}^bdisplaystylesumlimits_{y=c}^d[(x,y)==k]$
题解:下文中令$nleqslant m$先考虑求$sumlimits_{i=1}^nsumlimits_{j=1}^m[(i,j)==k]$
$$
egin{align*}
defdsum{displaystylesumlimits}
令f(p)&=dsum_{i=1}^ndsum_{j=1}^m[(i,j)==p]\
令F(p)&=dsum_{p|k}f(k)\
&=dsum_{p|k}dsum_{i=1}^ndsum_{j=1}^m[(i,j)==k]\
&=dsum_{i=1}^ndsum_{j=1}^m[p|(i,j)]\
&=leftlfloordfrac{n}{p}
ight
floorcdot leftlfloordfrac{m}{p}
ight
floor\
莫比乌&斯反演得:\
herefore f(p)&=dsum_{p|k}muBig(dfrac{k}{p}Big)F(k)\
&=dsum_{i=1}^nmu(i)leftlfloordfrac{n}{ip}
ight
floorcdotleftlfloordfrac{m}{ip}
ight
floor\
end{align*}\
令g(p)=dsum_{i=1}^pmu(i)\
然后容斥一下就好了\
$$
卡点:无
C++ Code:
#include <cstdio> #define maxn 50010 using namespace std; int miu[maxn], plist[maxn], ptot; bool isp[maxn]; void sieve(int n) { miu[1] = 1; for (int i = 2; i <= n; i++) { if (!isp[i]) { miu[i] = -1; plist[ptot++] = i; } for (int j = 0; j < ptot, i * plist[j] <= n; j++) { int tmp = i * plist[j]; isp[tmp] = true; if (i % plist[j] == 0) { miu[tmp] = 0; break; } miu[tmp] = -miu[i]; } } for (int i = 2; i <= n; i++) miu[i] += miu[i - 1]; } inline int min(int a, int b) {return a < b ? a : b;} int solve(int n, int m, int k) { n /= k, m /= k; int tmp = min(n, m); int ans = 0, l, r; for (l = 1; l <= tmp; l = r + 1) { r = min(n / (n / l), m / (m / l)); ans += (miu[r] - miu[l - 1]) * (n / l) * (m / l); } return ans; } int Tim, a, b, c, d, k; int main() { sieve(50000); scanf("%d", &Tim); while (Tim --> 0) { scanf("%d%d%d%d%d", &a, &b, &c, &d, &k); printf("%d ", solve(b, d, k) - solve(a - 1, d, k) - solve(b, c - 1, k) + solve(a - 1, c - 1, k)); } return 0; }