Codeforces Round #475 (Div. 2) C - Alternating Sum

Posted notnight

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Codeforces Round #475 (Div. 2) C - Alternating Sum相关的知识,希望对你有一定的参考价值。

等比数列求和一定要分类讨论!!!!!!!!!!!! 

 1 #include<bits/stdc++.h>
 2 #define LL long long
 3 #define fi first
 4 #define se second
 5 #define mk make_pair
 6 #define pii pair<int,int>
 7 #define ull unsigned long long
 8 using namespace std;
 9 
10 const int N=1e6+7;
11 const int M=100+7;
12 const int inf=0x3f3f3f3f;
13 const LL INF=0x3f3f3f3f3f3f3f3f;
14 const int mod=1e9 + 9;
15 
16 LL n, a, b, k;
17 char s[N];
18 
19 LL q_pow(LL a, LL b) {
20     LL ans = 1;
21     while(b) {
22         if(b & 1) ans = ans * a % mod;
23         a = a * a % mod; b >>= 1;
24     }
25     return ans;
26 }
27 
28 LL ivn(LL x) {
29     return q_pow(x, mod - 2);
30 }
31 
32 
33 int main() {
34 
35     scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
36     scanf("%s", s);
37     LL w = q_pow(b, k) * ivn(q_pow(a, k)) % mod;
38     w = (w + mod) % mod;
39 
40     LL ans = 0;
41     LL cnt = (n + 1) / k;
42 
43     for(LL i = 0; i < k; i++) {
44         LL ret1 = q_pow(a, n - i) * q_pow(b, i) % mod;
45         LL ret2 = (1 - q_pow(w, cnt) + mod) % mod;
46         LL ret3 = ivn((1 - w + mod) % mod);
47         LL ret = (((ret1 * ret2) % mod) * ret3) % mod;
48         if(w == 1) ret = (ret1 * cnt) % mod;
49         if(s[i] == -) ret = -ret;
50         ans = (ans + ret + mod) % mod;
51     }
52     printf("%lld\n", ans);
53     return 0;
54 }
55 /*
56 */

 

以上是关于Codeforces Round #475 (Div. 2) C - Alternating Sum的主要内容,如果未能解决你的问题,请参考以下文章

Codeforces Round #475 (Div. 2) C - Alternating Sum

Codeforces Round #475 (Div. 2) C - Alternating Sum

Codeforces Round #475 Div. 2 A B C D

Codeforces Round #475 (Div. 2) D. Destruction of a Tree

Codeforces Round #436 E. Fire(背包dp+输出路径)

[ACM]Codeforces Round #534 (Div. 2)