POJ-3090-Visible Lattice Points（欧拉函数）

最新推荐文章于 2021-04-12 17:56:37 发布

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最新推荐文章于 2021-04-12 17:56:37 发布

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分类专栏：算法练习数论文章标签：数论

本文链接：https://blog.youkuaiyun.com/qq_43461168/article/details/108327339

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本文探讨了一种算法问题，即在第一象限中计算不可见格点的数量。通过使用欧拉函数和前缀和技巧，文章提供了一个有效的解决方案，用于找出在给定范围内不受其他格点遮挡的点。

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Visible Lattice Points

A lattice point (x, y) in the first quadrant (x and y are integers greater than or equal to 0), other than the origin, is visible from the origin if the line from (0, 0) to (x, y) does not pass through any other lattice point. For example, the point (4, 2) is not visible since the line from the origin passes through (2, 1). The figure below shows the points (x, y) with 0 ≤ x, y ≤ 5 with lines from the origin to the visible points.

在这里插入图片描述

Write a program which, given a value for the size, N, computes the number of visible points (x, y) with 0 ≤ x, y ≤ N.

Input

The first line of input contains a single integer C (1 ≤ C ≤ 1000) which is the number of datasets that follow.

Each dataset consists of a single line of input containing a single integer N (1 ≤ N ≤ 1000), which is the size.

Output

For each dataset, there is to be one line of output consisting of: the dataset number starting at 1, a single space, the size, a single space and the number of visible points for that size.

Sample Input

Sample Output

1 2 5
2 4 13
3 5 21
4 231 32549

解题思路：

求不被遮挡的点的个数。
首先可以发现这个图是对称的。
那么只需要算右下部分的点*2 + 1 就可以得出答案。

很容易发现只有当gcd(x,y) == 1时，这个点才不会被遮挡。
因为如果 gcd(x,y) != 1 令 gcd(x,y) = d

那么 gcd(x/d,y/d) == 1 这个点是可以被看到的，所以(x,y)和他成倍数关系那么必然被遮挡。

接下来不就变成欧拉函数水题了嘛。可以求前缀和加速。但是n不大无所谓。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n,m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n,m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sc(n) scanf("%c",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,a,n) for(int i=n;i>=a;i--)
#define mem(a,n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define mod(x) ((x)%MOD)
#define gcd(a,b) __gcd(a,b)
#define lowbit(x) (x&-x)
#define pii map<int,int>
#define mk make_pair
#define rtl rt<<1
#define rtr rt<<1|1
#define Max(x,y) (x)>(y)?(x):(y)
#define int long long


typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const ll mod = 10007;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
//const int inf = 0x3f3f3f3f;
inline int read(){int ret = 0, sgn = 1;char ch = getchar();
while(ch < '0' || ch > '9'){if(ch == '-')sgn = -1;ch = getchar();}
while (ch >= '0' && ch <= '9'){ret = ret*10 + ch - '0';ch = getchar();}
return ret*sgn;}
inline void Out(int a){if(a>9) Out(a/10);putchar(a%10+'0');}
int qpow(int m, int k, int mod){int res=1%mod,t=m%mod;while(k){if(k&1)res=res*t%mod;t=t*t%mod;k>>=1;}return res;}
ll gcd(ll a,ll b){if(b > a) swap(a,b); return b==0?a : gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll inv(ll x,ll mod){return qpow(x,mod-2,mod)%mod;}

const int N = 1e6+15;
int primes[N], euler[N], pcnt;
bool st[N];
void getEulers(int n)
{ //欧拉筛的扩展
	euler[1] = 1;
    pcnt = 0;
	for (int i=2; i <= n; i++ ){
		if ( !st[i] ){
			primes[pcnt++] = i;
			euler[i] = i-1;
		}
		for ( int j=0; primes[j] <= n/i; j++ ){
			st[ primes[j]*i ] = true;
			if ( i % primes[j] == 0 ) {
				euler[ i*primes[j] ] = euler[i] * primes[j];
				break;
			}
			euler[ i*primes[j] ] = euler[i] * ( primes[j]-1 );
		}
	}
}


signed main()
{
    getEulers(N);
    int t = 1,cas = 1;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int res = 0;
        for(int i = 1 ; i <= n ; i ++)
            res += euler[i];
        cout<<cas++<<" "<<n<<" "<<res*2+1<<endl;

    }
}