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）

题目链接：
BZOJ 2301 Problem B
题意:
区间 x∈[a,b],y∈[c,d]满足gcd(x,y)=k的(x,y)的有序对数
分析：
莫比乌斯反演+优化。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int MAX_N = 50010;
typedef long long ll;

bitset 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;
}