2277: [Poi2011]Strongbox
Time Limit: 60 Sec Memory Limit: 32 MBSubmit: 498 Solved: 218
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Description
Byteasar is a famous safe-cracker, who renounced his criminal activity and got into testing and certifying anti-burglary devices. He has just received a new kind of strongbox for tests: a combinatorial safe. A combinatorial safe is something different from a combination safe, even though it is opened with a rotary dial. The dial can be set in different positions, numbered from 0 to n-1. Setting the dial in some of these positions opens the safe, while in others it does not. And here is the combinatorial property, from which the name comes from: if x and y are opening positions, then so is (x+y) mod n too; note that is holds for x=y as well.
Byteasar tried k different positions of the dial: m1,m2….mk. The positions M1,M 2….Mk-1 did not open the safe, only the last position Mk did. Byteasar is already tired from checking these K positions and has thus absolutely no intention of trying the remaining ones. He would like to know however, based on what he already knows about the positions he tried, what is the maximum possible number of positions that open the safe. Help him by writing an appropriate program!
有一个密码箱，0到n-1中的某些整数是它的密码。
且满足，如果a和b都是它的密码，那么(a+b)%n也是它的密码(a,b可以相等)
某人试了k次密码，前k-1次都失败了，最后一次成功了。
问：该密码箱最多有多少不同的密码。
Input
The first line of the standard input gives two integers N and k, separated by a single space, (1<=K<=250000,k<=N<=10^14), The second line holds K different integers, also separated by single spaces, m1,m2….mk, 0<=Mi<N. You can assume that the input data correspond to a certain combinatorial safe that complies with the description above.
In tests worth approximately 70% of the points it holds that k<=1000. In some of those tests, worth approximately 20% of the points, the following conditions hold in addition: N< 10 ^8 and K<=100.
第一行n，k
下面一行k个整数，表示每次试的密码
保证存在合法解
1<=k<=250000 k<=n<=10^14
Output
Your program should print out to the first and only line of the standard output a single integer: the maximum number of the dial's positions that can open the safe.
一行，表示结果
Sample Input
42 5
28 31 10 38 24
28 31 10 38 24
Sample Output
14
分析：比较难的一道数学题.有两个结论：1.如果x是密码，那么gcd(x,n)也是密码. 2.如果x,y是密码，那么gcd(x,y)也是密码.根据这两个结论就能很轻松地解决本题了.
先来证明第一个结论：构造二元一次不定方程x*k - n*c = gcd(x,n),这个方程是一定有解的，也就是说一定存在一个 k使得x*k%n = gcd(x,n).而x是密码，(x+x)%n也是密码，所以k*x%n也是密码，那么gcd(x,n)就是密码.x就是题目中告诉的a[k].
再来证明第二个结论：gcd(x,y) = a*x + b*y,a,b有可能小于0.根据结论一可以推出
(p*x + q*y)%n是密码(p,q ≥ 0). 由a*x + b*y ≡ gcd(x,y)(mod n),变形一下可以得到:
a*x + b*y ≡ a*x + b*y + p*n*x + q*n*y(mod n) --> (a + p*n) * x + (b + q*n) * y ≡ gcd(x,y)(mod n).根据假设，((a + p*n) * x + (b + q*n) * y) % n是密码(x,y系数大于0)，那么gcd(x,y)也是密码.
把所有密码写出来以后，可以发现是一个x,2x,3x......的形式，所以任务就是找到一个最小的x使得x整除gcd(a[k],n).同时这个x不能整除a[j](1 ≤ j ＜ k).那么x就是gcd(a[k],n)的因子，根号的时间处理出来然后进行判断.判断的话也有一个优化，如果y是密码，gcd(x,y)不是密码，那么x也不是密码，所以在判断的时候看一下所有的gcd(a[j],n)是否被当前的因子整除就行了.
#include <cstdio> #include <cmath> #include <cstring> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; ll n,k,a[250010],ans = n,cnt; bool check(ll x) { for (int i = 1; i <= cnt; i++) if (a[i] % x == 0) return false; return true; } ll gcd(ll a,ll b) { if (!b) return a; return gcd(b,a % b); } int main() { scanf("%lld%lld",&n,&k); for (int i = 1; i <= k; i++) { scanf("%lld",&a[i]); a[i] = gcd(a[i],n); } sort(a + 1,a + k); for (int i = 1; i < k; i++) if (a[i] != a[i - 1]) a[++cnt] = a[i]; for (ll i = 1; i <= sqrt(a[k]); i++) if (a[k] % i == 0) { if (check(i)) { ans = n / i; break; } else if (check(a[k] / i)) ans = n / a[k] * i; } printf("%lld ",ans); return 0; }