x * 2m = l (mod n + 1),故

x = l * (2m)-1 (mod n + 1)

只需要求一下逆元什么的就做完了,注意乘法要用"快速加"的方法。。。否则会爆long long

 

 1 /**************************************************************
 2     Problem: 1965
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:0 ms
 7     Memory:804 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11  
12 using namespace std;
13 typedef long long ll;
14  
15 ll n, m, l, p;
16  
17 ll mul(ll x, ll y, ll mod) {
18   ll res = 0;
19   while (y) {
20     if (y & 1) (res += x) %= mod;
21     (x <<= 1) %= mod;
22     y >>= 1;
23   }
24   return res;
25 }
26  
27 ll pow(ll x, ll y, ll mod) {
28   ll res = 1;
29   while (y) {
30     if (y & 1) res = mul(res, x, mod);
31     x = mul(x, x, mod);
32     y >>= 1;
33   }
34   return res;
35 }
36  
37 int main() {
38   scanf("%lld%lld%lld", &n, &m, &l);
39   p = n / 2 + 1;
40   p = pow(p, m, n + 1);
41   printf("%lld\n", mul(p, l, n + 1));
42   return 0;
43 }
View Code

 

By Xs酱~ 转载请说明 博客地址:http://www.cnblogs.com/rausen
posted on 2015-02-01 23:21  Xs酱~  阅读(141)  评论(0编辑  收藏  举报