Alternating Sum

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

不考虑正负的话，每两项之间之间公比为b/a,考虑正负，则把k段作为循环节，循环节育循环节之间公比为(b/a)^k,在把第一个k小节整体看作第一项，等比数列求和。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
#define ll long long
const int mod=1e9+;
ll a,b,k,n;
char s[];
ll quick_pow(ll x,ll y)
{
    ll ans=;
    while(y)
    {
        if(y&) ans=ans*x%mod;
        x=x*x%mod;
        y>>=;
    }
    return ans%mod;
}
int main()
{
    scanf("%lld%lld%lld%lld",&n,&a,&b,&k);
    scanf("%s",s);
    ll ans=,r=quick_pow(a,mod-)*b%mod,top=,pos;
    for(int i=;i<k;i++)
        ans=(ans+(s[i]=='+'?1ll:-1ll)*quick_pow(a,n-i)%mod*quick_pow(b,i)%mod)%mod;
    n=(n+)/k;
    ll q=quick_pow(r,k);
    if(q==) pos=ans*n%mod;
    else pos=ans*(quick_pow(q,n)-)%mod*quick_pow(q-,mod-)%mod;
    printf("%lld\n",(pos+mod)%mod);
    return ;
}