先二分找到这个非平方数，然后对一个一个区间进行求和的预处理。要注意边界问题。

代码如下：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;typedef long long Int;Int N, num;long long int sum[1000005];Int bsearch(Int l, Int r) {    Int mid, left;    while (l <= r) {        mid = (l + r) >> 1;        left = mid - (int)floor(sqrt(double(mid)));        if (N > left) {            l = mid + 1;        }        else {            r = mid - 1;        }    }    return l;}void pre(){    int lim = 1 << 16;    for (int i = 1; i <= lim; ++i) {        sum[i] = sum[i-1]+1LL * (2*i+1) * i;    }}int main(){    Int T, pos, s;    long long int ans;    pre();    scanf("%I64d", &T);    while (T--) {        ans = 0;        scanf("%I64d", &N);        pos = bsearch(1, 1LL<<32);        s = (int)floor(sqrt(double (pos)))-1;        ans += (s+1)*(pos - (s+1)*(s+1)+1);        printf("%I64d %I64d\n", pos, ans + sum[s]);    }    return 0;}