Codeforces Round #191 (Div. 2) C. Magic Five 快速幂

C. Magic Five

链接：

http://codeforces.com/contest/327/problem/C

题解：

求等比为k的等比数列之和T[n]..

当n为偶数..T[n] = T[n/2] + pow(k,n/2) * T[n/2]

当n为奇数...T[n] = T[n/2] + pow(k,n/2) * T[n/2] + 等比数列第n个数的值

比如 1+2+4+8 = (1+2) + 4*(1+2)

1+2+4+8+16 = (1+2) + 4*(1+2) + 16

代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <string>
 8 #include <map>
 9 #include <set>
10 #include <stack>
11 #include <queue>
12 #include <functional>
13 using namespace std;
14 #define rep(i,a,n) for (int i=a;i<=n;i++)
15 #define per(i,a,n) for (int i=n;i>=a;i--)
16 #define pb push_back
17 #define mp make_pair
18 #define all(x) (x).begin(),(x).end()
19 #define fi first
20 #define se second
21 #define SZ(x) ((int)(x).size())
22 typedef vector<int> VI;
23 typedef long long ll;
24 typedef pair<int, int> PII;
25 const ll mod = 1000000007;
26 // head
27 
28 char s[100004];
29 ll m;
30 ll POW(ll a, ll k)
31 {
32     ll x, ans = 1;
33     x = a;
34     while (k) {
35         if (k % 2) ans = (ans*x) % mod;
36         x = (x*x) % mod;
37         k /= 2;
38     }
39     return ans;
40 }
41 ll T(ll n, ll t)
42 {
43     if (n == 1) return t;
44     ll data = T(n / 2, t);
45     data = (data + data*POW(m, n / 2)) % mod;
46     if (n % 2) data = (data + POW(m, (n - 1))*t) % mod;
47     return data;
48 }
49 int main()
50 {
51     int k, i;
52     ll ans, x, len;
53     scanf("%s", s);
54     scanf("%d", &k);
55     len = strlen(s);
56     m = POW(2, len);
57     ans = 0;
58     x = 1;
59     for (i = 0; i<len; i++){
60         if (s[i] == '0' || s[i] == '5')
61             ans = (ans + x) % mod;
62         x = (x * 2) % mod;
63     }
64     ans = T(k, ans);
65     printf("%I64d
", ans);
66     return 0;
67 }