HDU

题目：给你一个m*n的网格，从(1,1)开始，你站在(0,0)，问你能看见多少个点

如果一些点在同一条直线上，你只能看见离你最近的那个点

思路：你只能看见gcd(r,c)=1的点，枚举r，算出有多少个c

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3fconst int maxn=1e5+50;int prime[maxn][20],tol[maxn];void Init(int n){    mm(tol,0);    for(int i=2;i<=n;i++){        if(tol[i])            continue;        for(int j=1;j*i<=n;j++)            prime[j*i][tol[j*i]++]=i;    }}void dfs(int i,int x,int cnt,int n,int up,LL &sum){    if(i==tol[n]){        if(cnt&1) sum-=up/x;        else sum+=up/x;        return;    }    dfs(i+1,x*prime[n][i],cnt+1,n,up,sum);    dfs(i+1,x,cnt,n,up,sum);}int main(){    int T,m,n;    Init(maxn-50);    scanf("%d",&T);    while(T--){        scanf("%d%d",&m,&n);        LL ans=n;        for(int i=2;i<=m;i++)            dfs(0,1,0,i,n,ans);        printf("%lld\n",ans);    }    return 0;}