- Div 3
There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Given first N elements of that sequence. You must determine amount of numbers in it that are divisible by 3.
Input
Input contains N (1<=N<=231 - 1).
Output
Write answer to the output.
Sample Input
4
Sample Output
2
题目类型：简单数学
算法分析：打表发现各位和能整除3的数是有规律的，直接按照规律求解即可(任意数模3的取值只有0、1和2)
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#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 100000 + 66; int main() { // ifstream cin ("aaa.txt"); // freopen ("aaa.txt", "r", stdin); long long n; cin >> n; if (n == 1) cout << "0" << endl; else if (n == 2) cout << "1" << endl; else { if (n % 3 == 0) cout << 2 * (n / 3) << endl; else { int tt = n / 3; if (n == 3 * tt + 1) cout << 2 * (n / 3) << endl; else if(n == 3 * tt + 2) cout << 2 * (n / 3) + 1 << endl; } } return 0; } |
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