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 ", mul(p, l, n + 1)); 42 return 0; 43 }