SPOJ BLAMOEBA - Super Amoeba

题目链接：http://www.spoj.com/problems/BLAMOEBA/

题目大意：有一种叫做阿米巴的虫子一天能分裂成M个虫子。现有X个这种虫子，从第一天开始进行分裂，每次分裂之后就取走Y个，问如果要第N天取走之后不再有虫子剩余，并且X尽可能小，那么X和Y应该为多少。2 < M < 100000 ,1 < N < 100000

解题思路：分析后可以得到 (mn-1)Y = mn(m-1)X. 那么关键就是对这个式子两边约分，当X,Y两个系数互素的时候就是答案了。如何约分呢？首先考虑 mn-1 与 m-1 的关系。

m % (m-1) = 1 ----> mn % (m - 1) = 1 ----> (mn - 1) % (m - 1) = 0. 所以，左右两边可以同时除以 m - 1。之后，由于 mn - 1 可以分解为 (m - 1) * (a1mn-1 + a2mn-2+...+am + 1) ，因此除以 m - 1 之后两者已经互素。

接下来就可以利用快速幂取模以及逆元来求X，Y对应的值了。

代码：

 1 ll n, m;
 2 
 3 ll ext_gcd(ll a, ll b, ll &d, ll &x, ll &y){
 4     if(b) {
 5         ext_gcd(b, a % b, d, y, x); y -= x * (a / b);
 6     }
 7     else{
 8         d = a; x = 1; y = 0;
 9     }
10 } 
11 ll pow_mod(ll a, ll b){
12     if(b == 0) return 1;
13     ll tmans = pow_mod(a, b / 2);
14     ll ans = tmans * tmans % mod;
15     if(b & 1) ans = a % mod * ans;
16     return ans % mod; 
17 } 
18 void solve(){
19     ll up = pow_mod(m, n);
20     ll x, y, d;
21     ext_gcd(m - 1, mod, d, x, y);
22     if(x < 0) x = (abs(x) / mod + 1) * mod + x;
23     ll dn = (up - 1) * x % mod;
24     printf("%lld %lld\n", dn, up);
25 }
26 int main(){
27     int t;
28     scanf("%d", &t);
29     while(t--){
30         scanf("%lld %lld", &m, &n);
31         solve();
32     }
33 }

题目：

BLAMOEBA - Super Amoeba

#math

 

Peter has an amoeba farm with pretty much unlimited amoebae. After years of research, Peter created a device to convert some amoebae to super amoebae. However, his device can only be used once. Every day, a super amoeba will split into M super amoebae (2 < M < 100000).

Now, Peter plan his amoeba selling business. Initially, Peter converts X amoebae to super amoebae (X > 1). Every day after the amoebae split, Peter will take Y super amoebae for sale (Y > 1). After N days, Peter want all of his amoebae to be completely sold out (1 < N < 100000). Since the energy needed to convert amoebae is quite massive, X must be as small as possible. Help peter plan his business!

Input

First line is T, number of test cases (T < 100000). Next T lines each contains M and N separated by space.

Output

For each case, output X and Y separated by space. Since X and Y can be very large, output them with modulo 1000000007.

Example

Input:
1
4 3

Output:
21 64

Explanation

Initially, Peter has 21 super amoebae.
After day 1, there are 4 x 21 - 64 = 20 super amoebae
After day 2, there are 4 x 20 - 64 = 16 super amoebae
After day 3, there are 4 x 16 - 64 = 0 super amoeba
All the super amoebae are sold out after the 3rd day just as planned.

 

posted @ 2017-08-22 08:36  EricJeffrey  阅读(259)  评论(0)  编辑  收藏  举报