# 2301 [HAOI2011]Problem b

2

2 5 1 5 1

1 5 1 5 2

14

3

### HINT

100%的数据满足：1≤n≤50000，1≤a≤b≤50000，1≤c≤d≤50000，1≤k≤50000

`c++

# include<bits/stdc++.h>

using namespace std;
const int maxn = 5e4 + 5;
int prime[maxn], mu[maxn];
bool not_prime[maxn];
int a, b, c, d, k, tot;

inline void prepare()
{
mu[1] = 1;
for(int i = 2; i < maxn; ++i){
if(!not_prime[i]){
prime[++tot] = i; mu[i] = -1;
}
for(int j = 1; prime[j] * i < maxn; ++j){
not_prime[prime[j] * i] = true;
if(i % prime[j] == 0){
mu[i * prime[j]] = 0;
break;
}
mu[i * prime[j]] = -mu[i];
}
}
for(int i = 1; i < maxn; ++i) mu[i] += mu[i - 1];
}

inline int workk(int n, int m)
{
if(n > m) swap(n, m);
int next, ret = 0;
for(int d = k; d <= n; d = next + k){
next = min(n / (n / d), m / (m / d));
next = (next / k) * k;
ret += (mu[next / k] - mu[d / k - 1]) * ((n / d) * (m / d));
}
return ret;
}

int main()
{
prepare();
int T; scanf("%d", &T);
while(T--){
scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);
printf("%d\n", workk(b, d) - workk(b, c - 1) - workk(a - 1, d) + workk(a - 1, c - 1));
}
return 0;
}

posted @ 2018-06-08 19:21  沛霖  阅读(76)  评论(0编辑  收藏