UVa 10214 Trees in a Wood. (数论-欧拉函数)

Posted 2020-08-31 dwtfukgv

题意：给定一个abs(x) <= a, abs(y) <= b,除了原点之外的整点各有一棵树，可以相互阻挡，求从原点可以看到多少棵树。

析：由于a < b，所以我们可以一列一列的统计，第 x 列可以看到的树的个数就是 0 < y <= b中gcd（x, y） = 1的y的个数。

然后就可以分别统计，时间复杂度为O(a*a)。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-5;
const int maxn = 2000 + 10;
const int mod = 1e6;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int phi[maxn];

void phi_table(){
  memset(phi, 0, sizeof phi);
  phi[1] = 1;
  for(int i = 2; i < maxn; ++i)  if(!phi[i])
    for(int j = i; j < maxn; j += i){
      if(!phi[j])  phi[j] = j;
      phi[j] = phi[j] / i * (i-1);
    }
}

int main(){
  phi_table();
  while(scanf("%d %d", &n, &m) == 2 && n){
    LL ans = 0;
    for(int i = 1; i <= n; ++i){
      ans += m / i * phi[i];
      int t = m % i;
      for(int j = 1; j <= t; ++j)  if(gcd(i, j) == 1)  ++ans;
    }
    ans = ans * 4LL + 4LL;
    LL tmp = (LL)n * m * 4LL + 2LL * (m+n);
    double res = (double)ans / ((double)tmp);
    printf("%.10f \n", res);
  }
  return 0;
}