int main() {
v[0] = 1;
v[1] = 1;
long long i, j;
for (i = 2; i * i < 1000; i++) {
if (!v[i]) {
for (j = i * i; j < 1000; j += i) {
v[j] = 1;
}
}
}
int index = 0;
for (i = 2; i < 1000; i++) {
if (!v[i]) {
primes[index++] = i;
}
}
int s;
int cases=1;
while(scanf("%d",&s)==1 && s)
{
int flag=0;
for(i=s;i>=0;i--)
{
long long returnvalue=sumDiv(i);
if(returnvalue==s)
{
flag=1;
printf("Case %d: %lld",cases,i);
break;
}
}
if(!flag)
{
printf("Case %d: -1",cases);
}
printf("\n");
}
return 0;
}
int main(int argc, char **argv)
{
long int res = sumDiv(3) + sumDiv(5) - sumDiv(15);
printf("Soma: %ld\n", res);
return 0;
}
int main( void )
{
unsigned long long a, b;
unsigned long long s;
a = 1;
s = 0;
while( a < 10000 )
{
b = sumDiv( a );
if( sumDiv( b ) == a && b > a )
s += ( a + b );
a += 1;
}
printf( "Answer: %llu\n", s );
return 0;
}
/*For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110;
therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.*/
int problem21(){
int aiDivisorSum[N];
for(int i=0;i<N;i++){
aiDivisorSum[i] = sumDiv(i);
}
int sum = 0;
for(int i=0;i<N;i++){
if(aiDivisorSum[i] < N && aiDivisorSum[aiDivisorSum[i]] < N)
if(aiDivisorSum[aiDivisorSum[i]] == i)
if(aiDivisorSum[i] != i)
sum += i;
}
return sum;
}
