題目連結
【分析】

• 就是求$\Sigma_{i=1}^n <n>_i$
• 與除法分塊相同，但是資料範圍更小，查詢次數更多
• 考慮遞推求解
• 設$f(n)=\Sigma_{i=1}^{n}<n>_i$
• 則$f(n-1)=\Sigma_{i-1}^{n-1}<n-1>_i$
• 不考慮$n \equiv 0(mod$ $i)$的情況，$f(n)=f(n-1)+(n-1)$
• 當$n \equiv 0(mod$ $i)$時，對$f(n)$的貢獻為$-(\sigma(n)-n)$
• 所以$f(n)=f(n-1)+(n-1)-(\sigma(n)-n)$
• $\sigma$可以用線性篩求出

【程式碼】

#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;

const int N = 1e7 + 10;

int T, n;
int p[N], pr;
LL sigma[N], f[N], low[N];

void sieve(int n) {
  sigma[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (low[i] == 0) {
      low[
 
i] = p[++pr] = i;
      sigma[i] = i + 1;
    }
    for (int j = 1, k = i * p[j]; j <= pr && k <= n; ++j, k = i * p[j]) {
      if (i % p[j] == 0) {
        low[k] = low[i] * p[j];
        if (low[k] == k)
          sigma[k] = (low[k] * p[j] - 1) / (p[j] - 1);
        else
          sigma[k] = sigma[i / low[i]] * sigma[low[i] * p[j]];
        break;
      } else {
        low[k] = p[j];
        sigma[k] = sigma[i] * sigma[p[j]];
      }
    }
  }
  for (int i = 1; i <= n; ++i)
    f[i] = f[i - 1] + (i - 1) - (sigma[i] - i);
}

int main() {
  sieve(10000000);
  scanf("%d", &T);
  for (int i = 1; i <= T; ++i) {
    scanf("%d", &n);
    printf("%lld\n", f[n]);
  }
  return 0;
}