
类似容斥的,用总的人数减去不会越狱的可能(m*Pow(m-1,n-1))即可。用快速幂
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#include<cstring> #include<algorithm> #include<cmath> #include<cstdlib> using namespace std; typedef long long LL; const LL Mod = 100003; LL n,m; LL (LL a,LL b){ a%=Mod;; LL Ans = 1; for (;b;b>>=1,a=a*a%Mod){ if (b&1) Ans=Ans*a%Mod; } return Ans; } int main(){ scanf("%lld%lld",&m,&n); printf("%lldn",((Pow(m,n)-m*Pow(m-1,n-1)) % Mod + Mod) % Mod); return 0; }
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