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，Sn = n+pn(n−1)/2
当k≠1时，an+p/(k−1) = k(an−1+p/(k-1))，等比数列，Sn = (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。
1 #include<iostream> 2 using namespace std; 3 4 #define LL long long 5 const LL MOD = 1e9+7; 6 LL n,k,p; 7 8 LL pow(LL a, LL x) 9 { 10 LL tmp=1; 11 while(x) { 12 if(x&1) 13 tmp = tmp*a%MOD; 14 a = a*a%MOD; 15 x>>=1; 16 } 17 return tmp; 18 } 19 20 int main() 21 { 22 ios::sync_with_stdio(false); 23 while(cin>>n>>k>>p) 24 { 25 if(k == 1) 26 cout<<(n+p*n*(n-1)/2)%MOD<<endl; 27 else { 28 LL t = pow(k,n); 29 LL ans = t*k+(p-1)*t-k*(n*p+1)+(n-1)*p+1; 30 ans = (ans%MOD+MOD)%MOD; 31 ans = ans*pow((k-1)*(k-1), MOD-2)%MOD; 32 cout<<ans<<endl; 33 } 34 } 35 36 return 0; 37 }
解法二：
杜教板子：https://www.cnblogs.com/liubilan/p/9520292.html
直接套杜教板子；
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <algorithm> 5 #include <vector> 6 #include <string> 7 #include <map> 8 #include <set> 9 #include <cassert> 10 #include<bits/stdc++.h> 11 12 #define rep(i,a,n) for (ll i=a;i<n;i++) 13 #define per(i,a,n) for (ll i=n-1;i>=a;i--) 14 #define pb push_back 15 #define mp make_pair 16 #define all(x) (x).begin(),(x).end() 17 #define fi first 18 #define se second 19 #define SZ(x) ((ll )(x).size()) 20 using namespace std; 21 typedef long long ll; 22 typedef vector<ll > VI; 23 24 typedef pair<ll ,ll > PII; 25 const ll mod=1000000007; 26 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;} 27 // head 28 29 ll _,n; 30 namespace linear_seq { 31 const ll N=10010; 32 ll res[N],base[N],_c[N],_md[N]; 33 34 vector<ll > Md; 35 void mul(ll *a,ll *b,ll k) { 36 rep(i,0,k+k) _c[i]=0; 37 rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod; 38 for (ll i=k+k-1;i>=k;i--) if (_c[i]) 39 rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod; 40 rep(i,0,k) a[i]=_c[i]; 41 } 42 ll solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+... 43 // prll f("%d ",SZ(b)); 44 ll ans=0,pnt=0; 45 ll k=SZ(a); 46 assert(SZ(a)==SZ(b)); 47 rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1; 48 Md.clear(); 49 rep(i,0,k) if (_md[i]!=0) Md.push_back(i); 50 rep(i,0,k) res[i]=base[i]=0; 51 res[0]=1; 52 while ((1ll<<pnt)<=n) pnt++; 53 for (ll p=pnt;p>=0;p--) { 54 mul(res,res,k); 55 if ((n>>p)&1) { 56 for (ll i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0; 57 rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod; 58 } 59 } 60 rep(i,0,k) ans=(ans+res[i]*b[i])%mod; 61 if (ans<0) ans+=mod; 62 return ans; 63 } 64 VI BM(VI s) { 65 VI C(1,1),B(1,1); 66 ll L=0,m=1,b=1; 67 rep(n,0,SZ(s)) { 68 ll d=0; 69 rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod; 70 if (d==0) ++m; 71 else if (2*L<=n) { 72 VI T=C; 73 ll c=mod-d*powmod(b,mod-2)%mod; 74 while (SZ(C)<SZ(B)+m) C.pb(0); 75 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod; 76 L=n+1-L; B=T; b=d; m=1; 77 } else { 78 ll c=mod-d*powmod(b,mod-2)%mod; 79 while (SZ(C)<SZ(B)+m) C.pb(0); 80 rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod; 81 ++m; 82 } 83 } 84 return C; 85 } 86 ll gao(VI a,ll n) { 87 VI c=BM(a); 88 c.erase(c.begin()); 89 rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod; 90 return solve(n,c,VI(a.begin(),a.begin()+SZ(c))); 91 } 92 }; 93 94 int main() { 95 ll n, k, p; 96 cin>>n>>k>>p; 97 ll sum[120]; 98 99 ///求出前10项 100 sum[1]=1; 101 for(ll i=2;i<=10;i++){ 102 sum[i]=(sum[i-1]*k%mod+p)%mod; 103 } 104 for(ll i=2;i<=10;i++){ 105 sum[i]=(sum[i-1]+sum[i])%mod; 106 } 107 108 vector<ll >v; 109 for(ll i=1;i<=10;i++){ 110 v.push_back(sum[i]); 111 112 } 113 printf("%lld ",linear_seq::gao(v,n-1)); 114 115 }