[数学] BZOJ3142 [Hnoi2013]数列

题解：http://www.cnblogs.com/jianglangcaijin/archive/2013/08/13/3254314.html

先差分，每一个差分对答案的贡献

然后就很水啦

#include<cstdio>
#include<cstdlib>
#include<cmath>
using namespace std;
typedef long long ll;

inline char nc()
{
	static char buf[100000],*p1=buf,*p2=buf;
	if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
	return *p1++;
}

inline void read(ll &x)
{
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

ll n,m,K,P,Ans;

inline ll Pow(ll a,ll b){
	ll ret=1;
	for (;b;b>>=1,(a*=a)%=P)
		if (b&1)
			(ret*=a)%=P;
	return ret;
}

int main()
{
	freopen("t.in","r",stdin);
	freopen("t.out","w",stdout);
	read(n); read(K); read(m); read(P);
	Ans=m*(m+1)/2%P*Pow(m,K-2)%P*(K-1)%P;
	Ans=((n%P*Pow(m,K-1)%P-Ans)%P+P)%P;
	printf("%lld\n",Ans);
	return 0;
}