• [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;
    }
    
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  • 原文地址:https://www.cnblogs.com/zxyqzy/p/10382117.html
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