HDU 2841 visible trees)

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 2841

There is an N * m square. Starting from (), each vertex has a tree. A person stands at () and asks him how many trees he can see. When (0, 0) and another point are in a straight line, he can only see the nearest one.

The question is to find the number of Y/X types in the square of M * n, because two Y/X points can only see one. The number of types of Y/X is the number of (x, y) X and Y. Where (1 <= x <= m, 1 <= n <= Y ).

In this way, the previous question is similar, and the number of mutual quality numbers with I in a range [1, m] is obtained. Here, 1 <= I <= n. First, find out which is not mutually qualitative with I, and then reject the I factorization quality factor.

# Include <stdio. h> # include <iostream> # include <map> # include <set> # include <bitset> # include <list> # include <stack> # include <vector> # include <math. h> # include <string. h> # include <queue> # include <string> # include <stdlib. h >#include <algorithm> # define ll _ int64 // # define ll long # define EPS 1e-9 # define PI ACOs (-1.0) using namespace STD; const ll INF = 1 <30; const int maxn = 100010; ll ans; int n, m; int FAC [maxn]; int prime [maxn]; int faccnt; void getprime () {bool flag [maxn]; memset (flag, false, sizeof (FLAG); prime [0] = 0; For (INT I = 2; I <maxn; I ++) {If (flag [I] = false) prime [++ prime [0] = I; for (Int J = 1; j <= prime [0] & I * prime [J] <maxn; j ++) {flag [I * prime [J] = true; if (I % prime [J] = 0) Break ;}} void getfac (INT num) {int TMP = num; faccnt = 0; For (INT I = 1; I <= prime [0] & prime [I] * prime [I] <= TMP; I ++) {If (TMP % prime [I] = 0) {FAC [faccnt ++] = prime [I]; while (TMP % prime [I] = 0) tmp/= prime [I];} if (TMP = 1) break;} If (TMP> 1) FAC [faccnt ++] = TMP;} ll solve (INT m) {// queue array int que [110]; int L = 0; que [L ++] = 1; for (INT I = 0; I <faccnt; I ++) {int K = L; For (Int J = 0; j <K; j ++) que [L ++] = FAC [I] * que [J] * (-1);} ll ANW = 0; For (INT I = 0; I <L; I ++) ANW + = m/que [I]; return ANW;} int main () {int test; scanf ("% d", & test ); getprime (); For (INT item = 1; item <= test; item ++) {scanf ("% d", & N, & M ); if (n> m) Swap (n, m); ans = 0; For (INT I = 1; I <= N; I ++) {getfac (I ); ans + = solve (m);} printf ("% i64d \ n", ANS);} return 0 ;}

HDU 2841 visible trees)