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
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; }