int main()
{
seive();
int n;
while(scanf("%d",&n)!=EOF)
{
if(n==1)printf("%d\n",n);
else
{
printf("1");
if(a[n]!=0)
{
while(n%a[n]==0)
{
printf(" x %d",a[n]);
if(a[n]==0)
break;
n=n/a[n];
if(a[n]==0)
break;
}
}
printf(" x %d\n",n);
}
}
return 0;
}
/*
*素性测试，对sqrt(n)内的素数进行测试即可，预处理求出sqrt(n)中的素数，
*假设该范围内素数的个数为s，那么复杂度降为O(s)。
*/
bool is_prime(int x) {
int p = seive (sqrt (x + 0.5));
for (int i=1; i<=p; ++i) {
if (x % prime[i] == 0) {
return false;
}
}
return x != 1;
}
int main()
{
int t, n;
seive(MAX_NUM);
scanf("%d", &t);
while (t--){
scanf("%d", &n);
printf("%d\n", euler_totient(n));
}
return 0;
}
int main()
{
int num;
seive();
while(1)
{
scanf("%d",&num);
if(num == 0)
break;
calcfactors(num);
}
return 0;
}
int main()
{
long long n,i,max,j;
seive(10000100);
while(1)
{
scanf("%lld",&n);
if(n==0)
break;
j=0;
max=0;
if(n==1||n==-1)
{
printf("-1\n");
continue;
}
if(n<0)
n=-n;
for(i=0;i<c&&n>=a[i];)
{
if(n%a[i]==0)
{
n=n/a[i];
if(a[i]>max)
{
max=a[i];
j++;
}
}
else
{
i++;
}
}
if(n!=1&&j==0)
{
printf("-1\n");
continue;
}
if(n!=1)
{
printf("%lld\n",n);
continue;
}
if(j==1)
printf("-1\n");
else
printf("%lld\n",max);
}
return 0;
}
int main()
{
seive();
pre();
while(scanf("%ld",&n)!=EOF)
{
long cnt=0;
for(i=1;i<=n;i++)
{
scanf("%ld",&a);
if(prime[a]==1)
cnt++;
}
printf("%ld\n",cnt);
}
return 0;
}
int main()
{
int num,i;
for(i=0;i<n;i++)
cum[i] = 0;
seive();
while(1)
{
if(scanf("%d",&num)==EOF)
break;
if(num == 0)
break;
calcfactors(num);
}
return 0;
}
main()
{
int tc,t1,t2;
seive(1002001);
scanf("%d",&tc);
while(tc--)
{
scanf("%d %d",&t1,&t2);
if(t1==1 && t2==1)printf("0\n");
else
{
while(!prime[t1])t1++;
while(!prime[t2])t2--;
printf("%d\n",prime[t2]-prime[t1]+1);
}
}
}
int main()
{
int i,t;
seive();
ar[0]=0;
ar[1]=0;
//f();
for(i=2;i<=max;i++)
ar[i]=ar[i-1]+aa[i];
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
printf("%lld\n",ar[n]);
}
return 0;
}
int main(void)
{
seive();
long long n, temp, count, i, root ;
while(scanf("%lld",&n)==1)
{
if(n==0||n==1)
{
printf("1\n");
continue;
}
else if(n<=30000000 && flag[n]==false)
{
printf("2\n");
continue;
}
count = 1 ;
root = sqrt(n);
for(i=0;prime[i]<=root;i++)
{
temp = 1;
for(;n%prime[i]==0;)
{
n/=prime[i];
temp++;
}
if(temp>1 && prime[i]>2 ) count*=temp;
root = sqrt(n);
}
if(n>1 && n!=2) count*=2;
printf("%lld\n",count);
}
return 0;
}
int main()
{
int n, *seivearr, i, j, count = 0, limit, maxi = -1, maxv, sum;
scanf("%d", &n);
seivearr = seive(n, 0);
cumulative[count++] = 0;
for (i = 2; i < n; i++)
if (!seivearr[i])
cumulative[count++] = i;
// Find the cumulative sum of all the primes below one million. At some point the sum will
// exceed one million. We need not go beyond that point. Fortunately, for this problem the
// limit sum exceeds a million within the first 600 primes or so
for (i = 1; i < count; i++)
{
cumulative[i] += cumulative[i - 1];
if (cumulative[i] >= ONE_MILLION)
break;
}
limit = i;
// The idea now is that the sum of N consecutive primes starting from i is given by
// cumulative[i + N] - cumulative[i - 1]. Just keep a running count of the max N
// Since the 'limit' is small (600 or so), the n^2 algorithm runs reasonable fast
for (i = 1; i < limit; i++)
{
for (j = i; j < limit; j++)
{
sum = cumulative[j] - cumulative[i - 1];
assert(sum < ONE_MILLION);
if (!seivearr[sum])
{ /* sum is prime */
if (maxi < (j - i + 1))
{
maxi = j - i + 1;
maxv = sum;
}
}
}
}
printf("%d, %d\n", maxi, maxv);
return 0;
}
int main()
{
seive();
/* for(i=0;i<1000;i++)
printf("%d ",prime[i]);
*/ int m,n,t,i,sq,flag;
//scanf("%d",&t);
t=readnum();
while(t--)
{
//scanf("%d%d",&m,&n);
m=readnum();
n=readnum();
if(m<=2)
{
printf("2\n");
m=3;
}
else if(m%2==0)
m=m+1;
for(i=m;i<=n;i+=2)
{
flag=0;
sq=sqrt(i);
for(j=1;prime[j]<=sq;j++)
{
if(i%prime[j]==0)
{
flag=1;
break;
}
}
if(flag==0)
fastWrite(i);
}
}
return 0;
}
int main()
{long int t,n,q,b,r,i,temp,x;
long long int a,m;
scanf("%ld",&t);
seive();
calc();
while(t--)
{scanf("%ld%llu%ld",&n,&m,&q);
while(q--)
{for(i=1;i<=n;i++)
total[i]=0;
scanf("%ld",&r);
b=n-r;
a=1;
if((n-r)>r)
{
temp=n-r;
b=r;
r=temp;
}
for(x=2;x<=n;x++)
{
total[x]=fact[n][x]-fact[r][x];
}
for(x=2;x<=n-r;x++)
total[x]=total[x]-fact[n-r][x];
for(i=2;i<=n;i++)
{
if(total[i])
{
a=(a*fastpower(i,total[i],m))%m;
}
}
printf("%llu\n",a);
}
}
return 0;
}
int main()
{
int n,t,tmp;
scanf("%d",&t);
seive();
for(i=0; i<z; i++)
{
for(j=0; j<z; j++)
{
tmp=prime[i]+2*prime[j];
if(tmp<=10000)
a[tmp]++;
else
break;
}
}
while(t--)
{
scanf("%d",&n);
printf("%d\n",a[n]);
}
return 0;
}
int main() {
int n, i;
scanf("%d", &n);
printf("%d\n", seive(n));
return 0;
}