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：

/**************************************************
 * Author     : xiaohao Z
 * Blog     : http://www.cnblogs.com/shu-xiaohao/
 * Last modified : 2014-03-30 13:19
 * Filename     : main.cpp
 * Description     : 
 * ************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<unsigned int,unsigned int> puu;
typedef pair<int, double> pid;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;

const int INF = 0x3f3f3f3f;
const double eps = 1E-6;
const int LEN = 1010;

int main()
{
//    freopen("in.txt", "r", stdin);

    int T;
    cin >> T;
    while(T--){
        int n;
        cin >> n;
        for(int i=2; i<n; i++){
            if(n%i == 1) {
                cout << i;
                break;
            }
        }
        for(int i=2; i<=n+1; i++){
            if(n%i == i-1) {
                cout << " " << i << endl;
                break;
            }
        }    
    }
    return 0;
}