[BZOJ2301][HAOI2011]Problem b（莫比乌斯反演）

把一个询问拆成44个：

∑i=1b∑j=1d[gcd(i,j)=k]∑i=1b∑j=1d[gcd(i,j)=k]

∑i=1a−1∑j=1d[gcd(i,j)=k]∑i=1a−1∑j=1d[gcd(i,j)=k]

∑i=1b∑j=1c−1[gcd(i,j)=k]∑i=1b∑j=1c−1[gcd(i,j)=k]

∑i=1a−1∑j=1c−1[gcd(i,j)=k]∑i=1a−1∑j=1c−1[gcd(i,j)=k]

然后就是莫比乌斯反演的经典模型∑ni=1∑mj=1[gcd(i,j)=t]∑i=1n∑j=1m[gcd(i,j)=t]了。

具体见：http://blog.csdn.net/xyz32768/article/details/79218250

代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
inline int read() {
int res = 0; bool bo = 0; char c;
while (((c = getchar()) < '0' || c > '9') && c != '-');
if (c == '-') bo = 1; else res = c - 48;
while ((c = getchar()) >= '0' && c <= '9')
res = (res << 3) + (res << 1) + (c - 48);
return bo ? ~res + 1 : res;
}
typedef long long ll;
const int N = 5e4;
int pri[N + 5], miu[N + 5], sum[N + 5], cnt;
bool is[N + 5];
void sieve() {
int i, j; miu[1] = 1; is[0] = is[1] = 1;
for (i = 2; i <= N; i++) {
if (!is[i]) pri[++cnt] = i, miu[i] = -1;
for (j = 1; j <= cnt; j++) {
if (1ll * i * pri[j] > N) break;
is[i * pri[j]] = 1;
if (i % pri[j] == 0) break;
else miu[i * pri[j]] = -miu[i];
}
}
for (i = 1; i <= N; i++) sum[i] = sum[i - 1] + miu[i];
}
ll solve(int n, int m) {
if (!n || !m) return 0ll;
int i, nxt; ll ans = 0;
for (i = 1; i <= min(n, m);) {
nxt = min(n / (n / i), m / (m / i));
ans += 1ll * (sum[nxt] - sum[i - 1]) * (n / i) * (m / i);
i = nxt + 1;
}
return ans;
}
int main() {
int T = read(), a, b, c, d, k; sieve();
while (T--) {
a = read(); b = read(); c = read();
d = read(); k = read();
printf("%lld\n", solve(b / k, d / k) - solve((a - 1) / k, d / k)
- solve(b / k, (c - 1) / k) + solve((a - 1) / k, (c - 1) / k));
}
return 0;
}