【牛客小白月赛6】 J 洋灰三角

题目地址：https://www.nowcoder.com/acm/contest/136/J

解法一：

　　推数学公式求前n项和；　

　　当k=1时，即为等差数列，S_n = n+pn(n−1)/2
　　当k≠1时，a_n+p/(k−1) = k(a_n−1+p/(k-1))，等比数列，S_n = (kⁿ⁺¹+(p−1)kⁿ−(np+1)k+(n−1)p+1) / ((k-1)*(k-1))

　　因为是除法取模，故除快速幂外还需逆元；

Knowledge Point：

　　除法取模逆元：https://www.cnblogs.com/ECJTUACM-873284962/p/6847672.html

　　费马小定理：若p是质数，且a、p互质，那么a^(p-1) mod p = 1。

#include<iostream>
using namespace std;
 
#define LL long long
const LL MOD = 1e9+7;
LL n,k,p;
 
LL pow(LL a, LL x)
{
    LL tmp=1;
    while(x) {
        if(x&1)
            tmp = tmp*a%MOD;
        a = a*a%MOD;
        x>>=1;
    }
    return tmp;
}
 
int main()
{
    ios::sync_with_stdio(false);
    while(cin>>n>>k>>p)
    {
        if(k == 1)
            cout<<(n+p*n*(n-1)/2)%MOD<<endl;
        else {
            LL t = pow(k,n);
            LL ans = t*k+(p-1)*t-k*(n*p+1)+(n-1)*p+1;
            ans = (ans%MOD+MOD)%MOD;
            ans = ans*pow((k-1)*(k-1), MOD-2)%MOD;
            cout<<ans<<endl;
        }
    }
     
    return 0;
}

解法二：

　　杜教板子：https://www.cnblogs.com/liubilan/p/9520292.html

　　直接套杜教板子；

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
#include<bits/stdc++.h>

#define rep(i,a,n) for (ll  i=a;i<n;i++)
#define per(i,a,n) for (ll  i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((ll )(x).size())
using namespace std;
typedef long long ll;
typedef vector<ll > VI;

typedef pair<ll ,ll > PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
// head

ll  _,n;
namespace linear_seq {
    const ll  N=10010;
    ll res[N],base[N],_c[N],_md[N];

    vector<ll > Md;
    void mul(ll *a,ll *b,ll  k) {
        rep(i,0,k+k) _c[i]=0;
        rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (ll  i=k+k-1;i>=k;i--) if (_c[i])
            rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        rep(i,0,k) a[i]=_c[i];
    }
    ll  solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
//        prll f("%d
",SZ(b));
        ll ans=0,pnt=0;
        ll  k=SZ(a);
        assert(SZ(a)==SZ(b));
        rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
        rep(i,0,k) res[i]=base[i]=0;
        res[0]=1;
        while ((1ll<<pnt)<=n) pnt++;
        for (ll  p=pnt;p>=0;p--) {
            mul(res,res,k);
            if ((n>>p)&1) {
                for (ll  i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
    VI BM(VI s) {
        VI C(1,1),B(1,1);
        ll  L=0,m=1,b=1;
        rep(n,0,SZ(s)) {
            ll d=0;
            rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==0) ++m;
            else if (2*L<=n) {
                VI T=C;
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+1-L; B=T; b=d; m=1;
            } else {
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    ll  gao(VI a,ll n) {
        VI c=BM(a);
        c.erase(c.begin());
        rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
    }
};

int  main() {
    ll   n, k, p;
    cin>>n>>k>>p;
    ll  sum[120];
    
    ///求出前10项
    sum[1]=1;
    for(ll  i=2;i<=10;i++){
        sum[i]=(sum[i-1]*k%mod+p)%mod;
    }
    for(ll  i=2;i<=10;i++){
        sum[i]=(sum[i-1]+sum[i])%mod;
    }

    vector<ll >v;
    for(ll  i=1;i<=10;i++){
        v.push_back(sum[i]);

    }
    printf("%lld
",linear_seq::gao(v,n-1));

}