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SP3871 GCDEX - GCD Extreme

题目让我们求
\[\sum_{i=1}^n\sum_{j=i+1}^{n}gcd(i,j)\]
\[\sum_{i=1}^n\sum_{j=1}^{i-1}gcd(i,j)\]
设\(g(n) = \sum_{i=1}^{n-1}gcd(i,n)\)
\[\sum_{i=1}^ng(i)\]
考虑如何快速求\(g\)
\[g(n)=\sum_{i=1}^{n-1}gcd(i,n)\]
设\(gcd(i,n)==d\)
换个方向枚举
\[g(n)=\sum_{d=1}^{n-1}d\sum_{i=1}^n[gcd(i,n) == d]\]
\[g(n)=\sum_{d=1}^{n-1}d\sum_{i=1}^n[gcd(i,n) == d]\]
\[g(n)=\sum_{d=1}^{n-1}d\sum_{i=1}^{\frac nd}[gcd(i,\frac nd) == 1]\]
\[g(n)=\sum_{d=1}^{n-1}d\phi(\frac nd)\]

然后筛一下g(n),直接前缀和O(1)询查
怎么筛：枚举因子d,然后直接暴力+d,这就是著名的调和级数

    rep(i , 1, N){
        for(int j = 2 * i;j <= N;j += i) 
            g[j] += i * phi[j / i];
    }

AC代码:

#include <iostream>
#include <cstdio>
#define rep(i , x, p) for(register int i = x;i <= p;++ i)
#define gc getchar()
#define pc putchar
#define ll long long
const int maxN = 1e7 + 7;

int num , prime[maxN], phi[maxN];
bool is_prime[maxN];
ll sum[maxN] , g[maxN];

inline void init() {
    int N = 1000000;
    phi[1] = 1;
    rep(i , 2, N) {
        if(!is_prime[i]) prime[++ num] = i , phi[i] = i - 1;
        for(register int j = 1;j <= num && i * prime[j] <= N;++ j) {
            is_prime[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 );
        }
    }
    rep(i , 1, N){
        for(int j = 2 * i;j <= N;j += i) 
            g[j] += i * phi[j / i];
    }
    rep(i , 1, N) sum[i] = sum[i - 1] + g[i];
}

int main() {
    int n;
    init();
    while(scanf("%d",&n) == 1 && n) {
        printf("%lld\n",sum[n]);    
    }
    return 0;
}

SP3871 GCDEX - GCD Extreme

标签：getch   快速   printf   prime   ret   long   etc   怎么   main   

原文地址：https://www.cnblogs.com/gzygzy/p/10122784.html