Pagini recente » Cod sursa (job #1468692) | Cod sursa (job #2232394) | Cod sursa (job #2740895) | Cod sursa (job #180294) | Cod sursa (job #1614690)
#include <stdio.h>
#include <stdlib.h>
//merci pauldb de solutie :)
using namespace std;
long long phi[100001], n;
int main()
{
int i, j;
long long x;
FILE *dinuin, *dinuout;
dinuin = fopen("sum.in", "r");
dinuout = fopen("sum.out", "w");
fscanf(dinuin, "%lld", &n);
for (i = 1; i <= 100000; i++) phi[i] = i;
for (i = 2; i <= 100000; i++)
if (phi[i] == i)
for (j = i; j <= 100000; j += i) phi[j] = phi[j] / i * (i - 1);
for (i = n; i > 0; i--) {
fscanf(dinuin, "%lld", &x);
fprintf(dinuout, "%lld\n", phi[x] * 2 * x);
}
return 0;
}