2018 Multi-University Training Contest 10 - Count

数论，gcd

gcd(i+j, i-j) = gcd(2*i, i+j)

把i+j看出一个整体，得到范围为[i+1...2*i-1]

再看phi[2i]的意义，即[1...2i-1]上与2*i互质的数的个数

i+j的范围刚好是其一半，又引理可知欧拉函数在这个范围内均分成两半。

所以当i为奇数时，2和i互质，根据积性函数的性质，phi[2i] = phi[2] phi[i] = phi[i]

当i为偶数时，由欧拉函数的性质，phi[2i] = 2phi[i]

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();
    return w ? -ret : ret;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template <typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template <typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 20000005;
int _, n, tot, phi[N], prime[N];
ll a[N];
bool vis[N];

void calc(){
    phi[1] = 1;
    for(int i = 2; i < N; i ++){
        if(!vis[i]){
            prime[++tot] = i;
            phi[i] = i - 1;
        }
        for(int j = 1; j <= tot && i * prime[j] < N; j ++){
            vis[i * prime[j]] = true;
            if(i % prime[j] == 0){
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            phi[i * prime[j]] = phi[i] * (prime[j] - 1);
        }
    }
    for(int i = 1; i < N; i ++){
        if(i & 1) a[i] = a[i - 1] + phi[i] / 2;
        else a[i] = a[i - 1] + phi[i];
    }
}

int main(){

    calc();
    for(_ = read(); _; _ --){
        n = read();
        printf("%lld\n", a[n]);
    }
    return 0;
}