[POI2007]ZAP-Queries 数学-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_40273481/article/details/87305855

本文详细解析了POI2007 ZAP-Queries数学题，探讨了如何通过编程自动化解决字节码破译过程中的数学查询问题，特别关注于求解特定条件下整数对的数量。

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[POI2007]ZAP-Queries 数学

题目描述

Byteasar the Cryptographer works on breaking the code of BSA (Byteotian Security Agency). He has alreadyfound out that whilst deciphering a message he will have to answer multiple queries of the form"for givenintegers

$1≤x≤a1\le x\le a1≤x≤a,1≤y≤b1\le y\le b1≤y≤b,gcd(x,y)=dgcd(x,y)=dgcd(x,y)=d, where gcd(x,y)gcd(x,y)gcd(x,y) is the greatest common divisor of xxx and yyy".$

Byteasar would like to automate his work, so he has asked for your help.

TaskWrite a programme which:

reads from the standard input a list of queries, which the Byteasar has to give answer to, calculates answers to the queries, writes the outcome to the standard output.

FGD正在破解一段密码，他需要回答很多类似的问题：对于给定的整数a,b和d，有多少正整数对x,y，满足x<=a，y<=b，并且gcd(x,y)=d。作为FGD的同学，FGD希望得到你的帮助。

输入输出格式

输入格式：

The first line of the standard input contains one integer $0001\le n\le 50\ 0001≤n≤50 000),denoting the number of queries.$

The following $0001\le d\le a,b\le 50\ 0001≤d≤a,b≤50 000), separated by single spaces.$

Each triplet denotes a single query.

输出格式：

Your programme should write

输入输出样例

输入样例#1： 复制

2
4 5 2
6 4 3

输出样例#1： 复制

3
2

#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 0x7fffffff
//#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 = 98765431;
#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 n;
int mu[maxn+10], vis[maxn+10];
int sum[maxn + 10];
void init() {
	for (int i = 1; i < maxn; i++)mu[i] = 1, vis[i] = 0;
	for (int i = 2; i < maxn; i++) {
		if (vis[i])continue;
		mu[i] = -1;
		for (int j = 2 * i; j < maxn; j += i) {
			vis[j] = 1;
			if ((j / i) % i == 0)mu[j] = 0;
			else mu[j] *= -1;
		}
	}
	for (int i = 1; i < maxn; i++)sum[i] = sum[i - 1] + mu[i];
}
int main()
{
	//	ios::sync_with_stdio(0);
	init();
	n = rd();
	while (n--) {
		int a = rd(), b = rd(), d = rd();
		ll ans = 0;
		for (int l = 1, r; l <= (min(a, b) / d); l = r + 1) {
			r = min((a / d) / (a / d / l), (b / d) / (b / d / l));
			ans += 1ll * (sum[r] - sum[l - 1])*(a / d / l)*(b / d / l);
		}
		cout << (ll)ans << endl;
	}
	return 0;
}