答案((n/2+1)^m*l)%(n+1)。
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; long long n,m,l; long long mul(long long a,long long b) { long long d=(long long)floor(a*(long double)b/(n+1)+0.5); long long ret=a*b-d*(n+1); if (ret<0) ret+=n+1; return ret; } long long f_pow(int a,int b) { int ans=1,base=a; while (b) { if (b&1) ans=mul(ans,base)%(n+1); base=mul(base,base)%(n+1); b>>=1; } return ans%(n+1); } int main() { scanf("%lld%lld%lld",&n,&m,&l); printf("%lld\n",mul(l,f_pow(n/2+1,m))%(n+1)); return 0; }
时间: 2024-10-15 02:21:03