codeforces 964 C. Alternating Sum(逆元、数学)

题目链接:http://codeforces.com/contest/964/problem/C

思路:这是一个以k为周期的函数,可以先求出第一个周期的和,而之后的周期等于是前一个周期乘上b/a的k次方,b/a的k次方,可以用逆元加费马小定理求出来。。。然后就是一个等比求和了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include<iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod int(1e9+7)
#define lc (d<<1)
#define rc (d<<1|1)
#define P pair<int,int>
#define pi acos(-1)
ll n,a,b,k;
string d;
ll gmod(ll a, ll b, ll c)
{
    long long res = 1;
    while(b>0)
    {
        if(b&1)
        {
            res=(res*a)%c;
        }
        a=(a*a)%c;
        b>>=1;
    }
    return res;
}
int main()
{
    cin.tie(0);
    cin>>n>>a>>b>>k;
    ll m=1e9+9,z=0,x,y,qw;
    cin>>d;
    FOR(i,0,k-1)
    {
        x=(gmod(a,n-i,m)%m+m)%m;
        y=(gmod(b,i,m)%m+m)%m;
        if(d[i]=='+')
        {
            z=(z%m+(x*y)%m)%m;
        }
        else
        {
            z=(z%m-(x*y)%m)%m;
        }
    }
    x=(gmod(gmod(a,m-2,m),k,m)%m+m)%m;
    y=(gmod(b,k,m)%m+m)%m;
    x*=y;
    x%=m;
    qw=(n+1)/k;
    ll s=0;
    if(x==1)  s=(z%m*qw%m)%m;
    else  s=(z%m*((gmod(x,qw,m)%m-1)%m*gmod(x-1,m-2,m)%m))%m;
    while(s<0) s+=m;
    s%=m;
    cout<<s<<endl;
    return 0;
}


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转载自blog.csdn.net/qq_40858062/article/details/79989991

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