POJ - 3090 gcd水题

Posted 2020-10-24 The Azure Arbitrator

大概题意就是求\(1 \le i,j \le n\)的\(gcd(i,j) = 1\)的个数+2(对于0的特判)
正解应该是欧拉函数或者高逼格的莫比乌斯反演
但数据实在太水直接打表算了

/*H E A D*/
bool GCD[1002][1002];
inline int gcd(int a,int b){return b?gcd(b,a%b):a;}
int main(){
    rep(i,1,1000) rep(j,1,1000) GCD[i][j]=bool(gcd(i,j)==1);
    int T=read(),kase=0;
    while(T--){
        int n=read();
        if(n==1){
            printf("%d %d 3\n",++kase,n);
            continue;
        }
        int ans=0;
        rep(i,1,n) rep(j,1,n) ans+=GCD[i][j];
        ans+=2;
        printf("%d %d %d\n",++kase,n,ans);
    }
    return 0;
}