BZOJ 2301 Problem B（x属于[a,b],y属于[c,d]满足gcd(x,y)=k的(x,y)的有序对数）

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <bitset>
using namespace std;
const int MAX_N = 50010;
typedef long long ll;

bitset<MAX_N> bs;
int prime_cnt, prime[MAX_N];
ll mu[MAX_N], sum[MAX_N];

void GetMu()
{
bs.set();
prime_cnt = 0;
memset(mu, 0, sizeof(mu));
mu[1] = 1;
for(int i = 2; i < MAX_N; ++i) {
if(bs[i] == 1){
mu[i] = -1;
prime[prime_cnt++] = i;
}
for(int j = 0; j < prime_cnt && i * prime[j] < MAX_N; ++j) {
bs[i* prime[j]] = 0;
if(i % prime[j]) {
mu[i * prime[j]] = -mu[i];
}else {
mu[i * prime[j]] = 0;
break;
}
}
}
for(int i = 1; i < MAX_N; ++i) {
sum[i] = sum[i - 1] + mu[i];
}
}

ll solve(int n, int m)//计算[1,n]和[1,m]区间有多少有序对(x, y)使得gcd(x,y) = 1
{
ll ans = 0;
int top = min(n, m), last;
for(int i = 1; i <= top; ++i) {
last = min(n / (n / i), m / (m / i));
ans += (ll) (n / i) * (m / i) * (sum[last] - sum[i - 1]);
i = last;
}
return ans;
}

int main()
{
GetMu();
int T, a, b, c, d, k;
scanf("%d", &T);
while(T--) {
scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);
if(k == 0){
printf("0\n");
continue;
}
ll ans = solve(b / k, d / k) + solve((a - 1) / k, (c - 1) / k) - solve((a - 1) / k, d / k) - solve((c - 1) / k, b / k);
printf("%lld\n", ans);
}
return 0;
}