SPOJ DIVISION

题目链接：http://www.spoj.com/problems/DIVISION/

题目大意：求0~2ⁿ-1 中有多少个数字能被3整除（包括0 和2ⁿ-1）。 n <= 2e18, 答案对 mod = 1e9 + 7取余。

解题思路：通过分析可以发现个数是 (2ⁿ / 3 + 1) % mod . 那么关键问题就是如何快速计算 2ⁿ / 3 % mod 。

由于n比较大，所以可以使用快速幂取模进行计算。然而，尽管a / b % c 可以通过逆元来计算，但前提是a整除b并且逆元存在。这里3关于mod 的逆元确实存在（互质），2ⁿ 却无法整除3，那么如何解决这个问题呢？

能够发现 2ⁿ% 3 = 1 或 2，并且当且仅当n为奇数为2，n为偶数为1。那么我们就可以将 2ⁿ / 3 % mod 转化成 2ⁿ - (n & 1? 2: 1) / 3 % mod.这样两者就能够整除了，从而也就可以采用快速幂取模计算了。

代码：

 1 ll n;
 2 ll x;
 3 
 4 ll ext_gcd(ll a, ll b, ll &d, ll &x, ll &y){
 5     if(b!= 0){
 6         ext_gcd(b, a % b, d, y, x);
 7         y -= x * (a / b);
 8     }
 9     else{
10         d = a; x = 1; y = 0;
11     }
12 }
13 void dowork(){
14     ll d, y;
15     ext_gcd(3, mod, d, x, y);
16     if(x < 0) x = x + (abs(x) / mod + 1) * mod;
17 }
18 ll pow_mod(ll a, ll b){
19     if(b == 0) return 1;
20     ll tmans = pow_mod(a, b / 2);
21     ll ans = tmans * tmans % mod;
22     if(b & 1) ans = a % mod * ans;
23     return ans;
24 }
25 void solve(){
26     ll ans = pow_mod(2, n);
27     ans = ans * x % mod;
28     ans = (ans - ((n & 1? 2: 1) * x) % mod) % mod; 
29     if(ans < 0) ans = (abs(ans) / mod + 1) * mod + ans;
30     ans = (ans + 1) % mod;
31     printf("%lld
", ans);
32 }
33 int main(){
34     dowork();
35     while(scanf("%lld", &n) != EOF){
36         solve();
37     }
38 }

题目：

DIVISION - Divisiblity by 3

#math #linear-algebra

Divisiblity by 3 rule is pretty simple rule:Given a number n sum the

digits of n and check if sum is divisible by 3.If divisible then n is

divisible by 3 else not divisible.

Seems pretty simple but what if we want to extend this rule in binary

representation!! Given a binary representation we can again find if it is

divisible by 3 or not.

Making it little bit interesting what if only length of binary

representation of a number is given say n.Can we find how many numbers

exist in decimal form such that when converted into binary form has n

length and is divisible by 3 ??

Divisiblity by 3 rule is pretty simple rule: Given a number sum the digits of number and check if sum is divisible by 3.If divisible then it is divisible by 3 else not divisible.Seems pretty simple but what if we want to extend this rule in binary representation!!

Given a binary representation we can again find if it is divisible by 3 or not. Making it little bit interesting what if only length of binary representation of a number is given say n.

Now can we find how many numbers exist in decimal form(base 10) such that when converted into binary(base 2) form has n length and is divisible by 3 ?? (1 <= n < 2*10^18)

Input

Length of binary form: n

output

Print in new line the answer modulo 1000000007.

Example

Input:
1

Output:
1

Explanation: For n=2 there are only 2 numbers divisible by 3 viz.0 (00) and 3 (11) and having length 2 in binary form.

NOTE:There are multiple testfiles containing many testcases each so read till EOF.

Warnings: Leading zeros are allowed in binary representation and slower languages might require fast i/o. Take care.