[HAOI2011]Problem b BZOJ2301 数学

题目描述

对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，且gcd(x,y) = k，gcd(x,y)函数为x和y的最大公约数。

输入输出格式

输入格式：

第一行一个整数n，接下来n行每行五个整数，分别表示a、b、c、d、k

输出格式：

共n行，每行一个整数表示满足要求的数对(x,y)的个数

输入输出样例

输入样例#1： 复制

2
2 5 1 5 1
1 5 1 5 2

输出样例#1： 复制

14
3

说明

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

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 10000000005ll
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int a, b, c, d, K;
int mu[maxn], vis[maxn], sum[maxn + 10];

void init() {
	for (int i = 1; i <= 50004; i++)mu[i] = 1, vis[i] = 0;
	for (int i = 2; i <= 50004; i++) {
		if (vis[i])continue;
		mu[i] = -1;
		for (int j = 2 * i; j <= 50004; j += i) {
			vis[j] = 1;
			if ((j / i) % i == 0)mu[j] = 0;
			else mu[j] *= -1;
		}
	}
	for (int i = 1; i <= 50004; i++)sum[i] = sum[i - 1] + mu[i];
}
int main()
{
	//	ios::sync_with_stdio(0);
	init();
	int  T = rd();
	while (T--) {
		cin >> a >> b >> c >> d >> K;
		ll ans1 = 0, ans2 = 0, ans3 = 0, ans4 = 0;
		for (int l = 1, r; l <= (min(b, d) / K); l = r + 1) {
			r = min((b / K) / (b / K / l), (d / K) / (d / K / l));
			ans1 += 1ll * (sum[r] - sum[l - 1])*(b / K / l)*(d / K / l);
		}
		for (int l = 1, r; l <= (min(a - 1, c - 1) / K); l = r + 1) {
			r = min((a - 1) / K / ((a - 1) / K / l), (c - 1) / K / ((c - 1) / K / l));
			ans2 += 1ll * (sum[r] - sum[l - 1])*((a - 1) / K / l)*((c - 1) / K / l);
		}
		for (int l = 1, r; l <= (min(a - 1, d) / K); l = r + 1) {
			r = min((a - 1) / K / ((a - 1) / K / l), (d) / K / ((d) / K / l));
			ans3 += 1ll * (sum[r] - sum[l - 1])*((a - 1) / K / l)*((d) / K / l);
		}
		for (int l = 1, r; l <= (min(b, c - 1) / K); l = r + 1) {
			r = min((b) / K / ((b) / K / l), (c - 1) / K / ((c - 1) / K / l));
			ans4 += 1ll * (sum[r] - sum[l - 1])*((b) / K / l)*((c - 1) / K / l);
		}
		cout << (ll)(ans1 + ans2 - ans3 - ans4) << endl;
	}
	return 0;
}