uva 11426 - GCD - Extreme (II)(数论)

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题目链接：uva 11426 - GCD - Extreme

题目大意：给出n，求∑i!=jngcd(i,j)

解题思路：f(n)=gcd(1,n)+gcd(2,n)+⋯+gcd(n−1,n)
S(n)=f(2)+f(3)+⋯+f(n)
S(n)=S(n−1)+f(n)
问题转化成怎么求f(n),对于一个n来说，枚举因子乘上个数即可。

#include <cstdio>#include <cstring>const int N = 4000000;typedef long long ll;int phi[N+5];ll s[N+5], f[N+5];void phi_table(int n) {    for (int i = 2; i <= n; i++)        phi[i] = 0;    phi[1] = 1;    for (int i = 2; i <= n; i++) {        if (!phi[i]) {            for (int j = i; j <= n; j += i) {                if (!phi[j])                    phi[j] = j;                phi[j] = phi[j] / i * (i-1);            }        }    }}int main () {    phi_table(N);    memset(f, 0, sizeof(f));    for (int i = 1; i <= N; i++) {        for (int j = i * 2; j <= N; j += i) {            f[j] += i * phi[j/i];        }    }    s[2] = f[2];    for (int i = 3; i <= N; i++)        s[i] = s[i-1] + f[i];    int n;    while (scanf("%d", &n) == 1 && n) {        printf("%lld\n", s[n]);    }    return 0;}

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