题解

不难看出，此题无疑是一道水爆了的题，缘由在于其数据范围$( n < 100 )$。 因此，我们仅需将$i$的三个数位一次性分离出来并直接判断是否三位均不等于$7$即可。 C++代码实现如下：

#include <queue>
#include <math.h>
#include <stack>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <iomanip>
#include <string.h>
#include <algorithm>
using namespace std;
#define int long long
#define float double
#define N 30000010
#define INF 0xc00000fd
#define scf scanf
#define ptf printf
#define gtc getchar
#define ptc putchar
queue<int> Q;
stack<int> S;
vector<int> V;
vector<int> travel(vector<int> A, vector<int> B, vector<int> U, vector<int> V, vector<int> W);
#ifndef SOMETHING_H
#define SOMETHING_H
#endif
inline int wei(int n){
	int cnt = 0;
	while(n > 0){
		n /= 10;
		cnt++;
	}
	return cnt;
}
inline bool isPrime(int n){
	if(n < 2)
		return 0;
	for(int i = 2; i * i <= n; ++i)
		if(n % i == 0)
			return 0;
	return 1;
}
inline int read(string n){
	int x = 0, f = 1;
	ptf("%s", n.c_str());
	char c = gtc();
	while(c < '0'  ||  c > '9'){
		if(c == '-')
			f = -1;
		c = gtc();
	}
	while(c >= '0'  &&  c <= '9'){
		x = x * 10 + c - 48;
		c = gtc();
	}
	return x * f;
}
inline float input(string n){
	float x = 0, f = 1, x2 = 0, cnt = 0, i = 0;
	ptf("%s", n.c_str());
	char c = gtc();
	while(c < '0'  ||  c > '9'){
		if(c == '-')
			f = -1;
		c = gtc();
	}
	while(c >= '0'  &&  c <= '9'){
		x = x * 10 + c - 48;
		c = gtc();
	}
	c = gtc();
	while(c >= '0'  &&  c <= '9'){
		x2 = x2 * 10 + c - 48;
		cnt++;
		c = gtc();
	}
	for(; i < cnt; i++)
		x2 /= 10.0;
	return (x + x2) * f;
}
inline void write(int n) {
    if(n < 0){
        ptc('-');
        n = -n;
    }
    if(n > 9)
		write(n / 10);
    ptc(n % 10 + '0');
	return;
}
inline void print(float n){
	ptf("%.12lf\n", n);
	return;
}
int n = read(""), sum;
inline void Main(){
	for(int i = 0; i < n; i++)
		if(i / 10 - 7  &&  i % 10 - 7  &&  i % 7)
			sum += i * i;
	write(sum);
	return;
}
signed main(signed argc, char **argv){
	Main();
	ptc('\n');
	return 0;
}