解题思路

令：

$$t=\left\lfloor\sqrt[3]x\right\rfloor$$

则：

$$\begin{aligned} \sum_{j=1}^x\left\lfloor j^{\frac{1}{3}}\right\rfloor & =\sum_{j=1}^x\left\lfloor\sqrt[3]j\right\rfloor \\ & =\sum_{j=1}^{t^3-1}\left\lfloor\sqrt[3]j\right\rfloor+\sum_{j=t^3}^x\left\lfloor\sqrt[3]j\right\rfloor \\ & =\sum_{i=1}^{t-1}\sum_{j=i^3}^{(i+1)^3-1}\left\lfloor\sqrt[3]j\right\rfloor+\sum_{j=t^3}^x\left\lfloor\sqrt[3]j\right\rfloor \\ & =\sum_{i=1}^{t-1}\sum_{j=i^3}^{(i+1)^3-1} i+\sum_{j=t^3}^x t \\ & =\left(\sum_{i=1}^{t-1}\left((i+1)^3-i^3\right)\cdot i\right )+(x-t^3+1)\cdot t \\ & =\left(\sum_{i=1}^{t-1} 3i^3+3i^2+i\right )+(x-t^3+1)\cdot t \\ & =\frac{3}{4} t^4-\frac{1}{2} t^3-\frac{1}{4} t^2-t^4+(x+1)\cdot t \\ & =(x+1)\cdot t-\frac{t^4+2t^3+t^2}{4} \end{aligned}$$

参考代码

261 Bcpp

#include <bits/stdc++.h>
using namespace std;

using ll=long long;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int q;
	cin>>q;
	while(q--)
	{
		ll x;
		cin>>x;
		ll t=cbrt(x+0.5);
		cout<<(x+1)*t-(t*t*t*t+2*t*t*t+t*t)/4<<'\n';
	}
	return 0;
}