1 solutions

  • 0
    @ 2025-11-5 19:21:05

    C++ :

    #include <bits/stdc++.h>
    using namespace std;
    
    int main() {
    	string s;
    	while (cin >> s) {
    		int sum = 0;
    		for (int i = 0; i < s.size(); i ++) {
    			sum += s[i] - '0';
    		}
    		
    		int res = sum % 9;
    		for (int i = 1; i <= s.size() - 1; i ++) {
    			res = res * 2 % 9;
    		}
    		cout << res << endl;
    	}
    
    	return 0;
    }
    
    
    /*
    n位数: 
    挑m个数字有C(n, m)种方法,共有C(n, m) * m个数字,
    对应n个数字重复C(n, m) * m / n遍,等于C(n - 1, m - 1)
    C(n - 1, 0) + C(n - 1, 1) + C(n - 1, 2) + ...... + C(n - 1, n - 1) = 2 ^ (n - 1)
    */
    
    • 1

    Information

    ID
    19277
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    (None)
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