poj3090 Visible Lattice Points

被机房oierA穿的一道题……

我最后还是靠着打表想出来了

其实就先沿着对角线分开 看一半

然后发现没有被覆盖的点纵坐标都是横坐标的因数

所以被覆盖的点就是欧拉函数

预处理一下前缀和 O(1000)

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 1005;
//head
int c;
int n;
int mindiv[N],prim[N],phi[N];
int sig[N];
void initphi(int n) {
    phi[0] = phi[1] = 1;
    mindiv[0] = mindiv[1] = 1;
    rep(i,2,n) {
    if(!mindiv[i]) prim[++prim[0]] = i,phi[i] = i-1;
    for(int j = 1;j <= prim[0] && i * prim[j] <= n;j++) {
        mindiv[i * prim[j]] = prim[j];
        if(i % prim[j] == 0) phi[i * prim[j]] = phi[i] * prim[j];
        else phi[i * prim[j]] = phi[i] * (prim[j]-1);
    }
    }
}

int main() {
    initphi(1001);
    rep(i,1,1000) sig[i] = sig[i-1] + phi[i];
    int T = read();
    rep(i,1,T) {
    int n = read();
    printf("%d %d %d
",i,n,(sig[n]<<1)+1);
    }
    return 0;
}