学步园

由题知：

g(n)=sigma{phi(n/d)*n/d},d|n

f(n)=(g(n)+1)/2

由于欧拉函数的积性，phi(n*m)=phi(n)*phi(m),gcd(n,m)=1

所以，n=pi(pi^ci)

故g(n)=g(pi(pi^ci))。

g(n)=pi(g(pi^ci)

g(pi^ci)=sigma(phi(pi^ci)pi^ci)

而对于phi(pi^ci)(pi-1)*(pi^(ci-1)

至此就这题就解决了！

我的代码：

超时的代码：

LL sumlcm(int n) {
int te = 0, temp = n;
    LL sum = 0;
if (n == 1)
return 1;
    te = (int) sqrt(n * 1.0);
    sum = (LL) n;
for (int i = 2; i <= te; i++) {
if (n % i == 0) {
            sum = sum / i * (i - 1);
while (n % i == 0) {
                n /= i;
            }
        }
    }
if (n > 1) {
        sum = sum / n * (n - 1);
    }
return sum * temp / 2;
}
void solve(int n) {
int i, te, temp;
    LL sum;
    te = (int) sqrt(n * 1.0);
for (i = 1, sum = 0; i <= te; i++) {
if (n % i == 0) {
            temp = n / i;
            sum += sumlcm(temp);
if (i != temp) {
                sum += sumlcm(i);
            }
        }
    }
    printf("%I64d\n", sum);
}

AC Code:

 

 

#include<stdio.h>
#include<string.h>
#include<math.h>
#define LL long long
#define nmax 44725
int flag[nmax], prime[nmax], plen;
void mkprime() {
    memset(flag, -1, sizeof(flag));
int i, j;
for (i = 2, plen = 0; i < nmax; i++) {
if (flag[i]) {
            prime[plen++] = i;
        }
for (j = 0; (j < plen) && (i * prime[j] < nmax); j++) {
            flag[i * prime[j]] = 0;
if (i % prime[j] == 0) {
break;
            }
        }
    }
}
int getPow(int a, int b) {
int res;
    res = 1;
while (b) {
if (b & 1) {
            res = res * a;
        }
        a = a * a;
        b >>= 1;
    }
return res;
}
int getpPhi(int p, int c) {
if (c == 0) {
return 1;
    }
return (p - 1) * getPow(p, c - 1);
}

LL calc(int p, int c) {
int i;
    LL temp, sum;
for (i = 0, sum = 0LL, temp = 1LL; i <= c; i++) {
        sum += temp * getpPhi(p, i);
        temp = temp * p;
    }
return sum;
}
void solve(int n) {
int i, te, cnt;
    LL res;
    te = (int) (sqrt(n * 1.0));
for (i = 0, res = 1LL; (i < plen) && (prime[i] <= te); i++) {
if (n % prime[i] == 0) {
            cnt = 0;
while (n % prime[i] == 0) {
                cnt++;
                n /= prime[i];
            }
            res = res * calc(prime[i], cnt);
        }
    }
if (n > 1) {
        res = res * calc(n, 1);
    }
    printf("%lld\n", (res + 1) / 2);
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
int n;
    mkprime();
while (scanf("%d", &n) != EOF) {
        solve(n);
    }
return 0;
}

 

 

#include<stdio.h>
#include<string.h>
#include<math.h>
#define LL long long
#define nmax 44725
int flag[nmax], prime[nmax], plen;
void mkprime() {
    memset(flag, -1, sizeof(flag));
int i, j;
for (i = 2, plen = 0; i < nmax; i++) {
if (flag[i]) {
            prime[plen++] = i;
        }
for (j = 0; (j < plen) && (i * prime[j] < nmax); j++) {
            flag[i * prime[j]] = 0;
if (i % prime[j] == 0) {
break;
            }
        }
    }
}
LL getPow(int a, int b) {
    LL res, temp;
    res = 1, temp = (LL) a;
while (b) {
if (b & 1) {
            res = res * temp;
        }
        temp = temp * temp;
        b >>= 1;
    }
return res;
}
LL calc(int p, int c) {
    LL res;
    res = getPow(p, 2 * c);
    res--;
    res = res / (p + 1);
    res = res * p;
    res++;
return res;
}
void solve(int n) {
int i, te, cnt;
    LL res;
    te = (int) (sqrt(n * 1.0));
for (i = 0, res = 1LL; (i < plen) && (prime[i] <= te); i++) {
if (n % prime[i] == 0) {
            cnt = 0;
while (n % prime[i] == 0) {
                cnt++;
                n /= prime[i];
            }
            res = res * calc(prime[i], cnt);
        }
    }
if (n > 1) {
        res = res * calc(n, 1);
    }
    printf("%lld\n", (res + 1) / 2);
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
int n;
    mkprime();
while (scanf("%d", &n) != EOF) {
        solve(n);
    }
return 0;
}