AcWing_1245 特别数的和(暴力枚举)

AcWing

1245. 特别数的和

真·暴力。枚举1~n的每一个数,判断是否有某一位是2/0/1/9.

#include <bits/stdc++.h>
using namespace std;
int n;
int main(){
    cin >> n;
    int res = 0;
    for(int i = 1; i <= n; i++){
        int t = i;
        while(t){
            int r = t % 10;
            if(r == 0 || r == 1 || r == 2 || r == 9){
                res += i;
                break;
            }
            t = t / 10;
        }
    }
    cout << res;
    return 0;
}