Codeforces 963A Alternating Sum (summation of proportional series + inverse element + fast power)

Question link: http://codeforces.com/problemset/problem/963/A

The main idea of ​​the title: It gives you n, a, b and a string of length k with only '+' and '-', ensuring that n+1 is divisible by k, allowing you to calculate .

Problem solving ideas:

Violence will definitely time out. We can first calculate the value of the segment 0~k-1 as a1. It can be found that if the value of each segment of length k is regarded as an element, they are proportional to each other. q=(b/a)^k,

Then you can directly use the equation series summation formula to find the answer. Yesterday, q was regarded as b/a, my brain is ah. . .

Note that when judging q==1, it cannot be achieved by judging a==b, but by judging (a/b)^k==1.

Code:

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<cctype>
 4 #include<cstring>
 5 #include<iostream>
 6 #include<algorithm>
 7 #include<vector>
 8 #include<queue>
 9 #include<set>
10 #include<map>
11 #include<stack>
12 #include<string>
13 #define lc(a) (a<<1)
14 #define rc(a) (a<<1|1)
15 #define MID(a,b) ((a+b)>>1)
16 #define fin(name)  freopen(name,"r",stdin)
17 #define fout(name) freopen(name,"w",stdout)
18 #define clr(arr,val) memset(arr,val,sizeof(arr))
19 #define _for(i,start,end) for(int i=start;i<=end;i++)
20 #define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
21 using namespace std;
22 typedef long long LL;
23 const LL MOD=1e9+9;
24 const double eps=1e-10;
25 
26 string str;
27 
28  LL fpow(LL x,LL n){
 29      LL res= 1 ;
 30      while (n> 0 ){
 31          if (n& 1 ) res=res*x%MOD;   // If the least significant bit of binary is 1, multiply On x^(2^i) 
32          x=x*x%MOD;                 //Squaring x and taking modulo 
33          n>>= 1 ;
 34      }
 35      return (res%MOD+MOD)% MOD;
 36  }
 37  
38 LL extend_gcd(LL a,LL b,LL &x,LL & y){
 39      if (! b){
 40          x= 1;
41         y=0;
42         return a;
43     }
44     LL gcd=extend_gcd(b,a%b,x,y);
45     LL t=x;
46     x=y;
47     y=t-(a/b)*x;
48     return gcd;
49 }
50 
51 LL NY(LL num){
52     LL x,y;
53     extend_gcd(num,MOD,x,y);
54     return (x%MOD+MOD)%MOD;
55 }
56 
57 int main(){
58     FAST_IO;
59     LL n,a,b,k;
60     cin>>n>>a>>b>>k;
61     cin>>str;
62     LL len=(n+1)/k;
63     LL sum=0;
64     for(int i=0;i<k;i++){
65         if(str[i]=='+')
66             sum=((sum+fpow(a,n-i)*fpow(b,i))%MOD+MOD)%MOD;
67         else
68             sum=((sum-fpow(a,n-i)*fpow(b,i))%MOD+MOD)%MOD;
69     }
70      LL ans;
 71      // Note that the ratio q is (b/a)^k not (b/a) 
72      LL q=fpow(NY(a),k)*fpow(b,k)% MOD;
 73      if (q!= 1 ){
 74          LL _q=NY(q- 1 );
 75          ans=(sum*(fpow(q,len)- 1 )%MOD*_q%MOD+MOD)% MOD;
 76      }
 77      else 
78          ans=sum*len% MOD;
 79      cout<<ans<< endl;
 80      return  0 ;
 81 }