bzoj 1965（快速幂+逆元）

传送门
数论收尾ing(ง •̀_•́)ง

/*    ans*(2^m)==L (mod n+1)    2 mod n+1 的逆元为 n/2+1(这一步用费马小定理求迷之WA)    ans=L*(n/2+1)^m (mod n+1)*/#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;inline ll fpow(ll a,ll b,ll MOD) {    ll ret=1;    while (b) {        if (b&1) ret=ret*a%MOD;        b>>=1,a=a*a%MOD;    }    return ret;}inline ll mult(ll x,ll y,ll MOD) {    x%=MOD,y%=MOD;    return ((x*y-(ll)(((long double)x*y+0.5)/MOD)*MOD)%MOD+MOD)%MOD;}int main() {    ll n,m,L;    cin>>n>>m>>L;    ll inv=(n>>1)+1;    cout<<mult(L,fpow(inv,m,n+1),n+1)<<endl;    return 0; }