factor[i]=i;
for(i=2;i<=n;i++){
if(i==factor[i])prime[pn++]=i;
for(j=0;j
if(i%prime[j]==0)
break;
}
}
}
int main()
{
int cases;
get_prime(N);
scanf("%d", &cases);
while(cases--)
{
int n, tmpn;
long long ans = 1;
scanf("%d", &n);
tmpn = n;
while (tmpn != factor[tmpn])
{
long long fac = factor[tmpn], mtp = fac;
while (tmpn % fac == 0)
{
mtp *= fac;
tmpn /= fac;
ans *= (1 - mtp) / (1 - fac);
}
if (tmpn > 1)
ans *= (1 + tmpn);
ans -= n;
printf("%lld\n", ans);
}
return 0;
}
SPOJ DIVSUM 另一种做法
[cpp]
#include
using namespace std;
int sum[500100];
int main()
{
int T,t,i,j,n;
scanf("%d",&T);
for(i=1;i<=500000;i++){
for(j=2;i*j<=500000;j++){//若j从1开始则会算上本身
sum[i*j]+=i;
}
}
for(t=1;t<=T;t++){
scanf("%d",&n);
printf("%d\n",sum[n]);
}
}