【牛客小白月赛6】 J 洋灰三角 - 快速幂&逆元&数学

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\n",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\n",linear_seq::gao(v,n-1));
114 
115 }

 

posted @ 2018-08-22 20:03  liubilan  阅读(167)  评论(0)  编辑  收藏  举报