Visible Trees

Visible Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

There are many trees forming a m * n grid, the grid starts from (1,1). Farmer Sherlock is standing at (0,0) point. He wonders how many trees he can see.

If two trees and Sherlock are in one line, Farmer Sherlock can only see the tree nearest to him.

 

Input

The first line contains one integer t, represents the number of test cases. Then there are multiple test cases. For each test case there is one line containing two integers m and n(1 ≤ m, n ≤ 100000)

 

Output

For each test case output one line represents the number of trees Farmer Sherlock can see.

 

Sample Input

2 1 1 2 3

 

Sample Output

1 5

分析：由题可知只有gcd（x，y）=1才能被看见；

　　　所以枚举其中一维坐标，容斥求出与之互质的另一维坐标即可；

　　　预处理每个数的素因子；

　　　或使用莫比乌斯函数，也是容斥求解；

代码1：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int maxn=1e5+10;
using namespace std;
inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline void umax(ll &p,ll q){if(p<q)p=q;}
inline void umin(ll &p,ll q){if(p>q)p=q;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t;
bool vis[maxn];
vi fac[maxn];
void init()
{
    for(int i=2;i<=maxn-10;i++)
    {
        if(vis[i])continue;
        for(int j=i;j<=maxn-10;j+=i)
        {
            vis[j]=true;
            fac[j].pb(i);
        }
    }
}
int gao(int x,int y)
{
    int ret=0,num=fac[x].size();
    for(int i=1;i<(1<<num);i++)
    {
        int now=1,cnt=0;
        for(int j=0;j<num;j++)
        {
            if(i&(1<<j))
            {
                cnt++;
                now*=fac[x][j];
            }
        }
        if(cnt&1)ret+=y/now;
        else ret-=y/now;
    }
    return y-ret;
}
int main()
{
    int i,j;
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&m,&n);
        ll ret=0;
        rep(i,1,m)
        {
            ret+=gao(i,n);
        }
        printf("%lld\n",ret);
    }
    return 0;
}

代码2：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int maxn=1e5+10;
using namespace std;
inline ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
inline ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline void umax(ll &p,ll q){if(p<q)p=q;}
inline void umin(ll &p,ll q){if(p>q)p=q;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,cnt,fac[maxn],ok[maxn];
int x,y;
bool vis[maxn];
void init()
{
    for(int i=2;i<=maxn-10;i++)
    {
        if(vis[i])continue;
        for(int j=i;j<=maxn-10;j+=i)
        {
            vis[j]=true;
            ++fac[j];
            if(j%((ll)i*i)==0)ok[j]=1;
        }
    }
}
int main()
{
    int i,j;
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&x,&y);
        ll ret=(ll)x*y;
        rep(i,2,maxn-10)
        {
            if(i>x||i>y)break;
            if(ok[i])continue;
            ret+=(ll)(fac[i]&1?-1:1)*(x/i)*(y/i);
        }
        printf("%lld\n",ret);
    }
    return 0;
}