Problem E Divisibility (11年珠海区域赛)
题意:找出b进制中n = ak * bk + ak-1 * bk-1 + ... + a1 * b + a0 除进 x 当且仅当 ak + ... + a0 除进 x, 和 n 除进 y 当且仅当 (a0 + a2 + ...) - (a1 + a3 + ...) 除进 y.
思路:x = 1(mod b) y = -1 (mod b)。想一下便知道了
代码如下:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-03-30 13:19 5 * Filename : main.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 const int LEN = 1010; 34 35 int main() 36 { 37 // freopen("in.txt", "r", stdin); 38 39 int T; 40 cin >> T; 41 while(T--){ 42 int n; 43 cin >> n; 44 for(int i=2; i<n; i++){ 45 if(n%i == 1) { 46 cout << i; 47 break; 48 } 49 } 50 for(int i=2; i<=n+1; i++){ 51 if(n%i == i-1) { 52 cout << " " << i << endl; 53 break; 54 } 55 } 56 } 57 return 0; 58 }
奔跑吧!少年!趁着你还年轻