Codeforces Round #191 (Div. 2) C. Magic Five rapid power of the proportional Series

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This question seems like the basic idea of POJ3233 is the same.

This question should be obtained using a fast power. If there are many proportional series of items, Division numbers should be included in the summation formula. Therefore, mod cannot be taken directly, and fast rice conversion should be performed.

For example, sum = 2 ^ 1 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + 2 ^ 5 + 2 ^ 6 + 2 ^ 7 .......

N items in total

This is the formula if n % 2 = 0 T (n) = T (n/2) + T (n/2) * 2 ^ (n/2 );

If n % 2 = 1 T (n) = T (n/2) + T (n/2) * 2 ^ (n/2) + 2 ^ n;

For this question, we first calculate the sum of the given cyclic bits as the basis, and then use the formula above the quick power to solve the problem,

I changed this question many times after the competition, especially the storage of _ int64 in its operation.

Below I will post my code ~~~ Write is a bit messy.

# Include <stdio. h> # include <iostream> # include <string. h> # include <algorithm> # include <math. h> # include <stdlib. h> using namespace std; const _ int64 mod = 1000000007 ;__ int64 all; int Pow (_ int64 n, int num) {// fast power num ^ n if (n = 0) return 1; if (n = 1) return num; _ int64 tmp = Pow (n/2, num) % mod; tmp = tmp * tmp % mod; if (n % 2 = 1) tmp = tmp * num % mod; return (int) tmp ;} __int64 len; int q; int fun (int n) {// the above-mentioned core formula if (n = 1) return al L; _ int64 tmp = fun (n/2); tmp = (tmp + tmp * Pow (n/2, q) % mod; if (n % 2 = 1) tmp = (tmp + Pow (n-1) * len, 2) * all % mod) % mod; return (int) tmp ;} char a [2000000]; int main () {int n, ans; while (scanf ("% s", )! = EOF) {cin> n; ans = 0; len = strlen (a); q = Pow (len, 2) % mod; all = 0; for (int I = 0; I <len; I ++) {if (a [I] = '0' | a [I] = '5 ') {all = (all + Pow (I, 2) % mod ;}// cout <I <"" <ans <endl ;} // cout <"all =" <all <endl; ans = fun (n); cout <ans <endl;} return 0 ;}