[Hoj 2901] calculation [binary theorem]

Most of the explanations are as follows:

Http://hi.baidu.com/timer/item/ebc8c7ef908085215b2d6476

Question:

(1 ^ B + 2 ^ B +... + A ^ B) %

1 ≤ A ≤ 1000000000, 1 ≤ B ≤ 1000000000

Ideas:

A and B have a large range. It is certainly not possible to directly calculate or directly use a rapid power modulo.

Note the question condition: B must be an odd number.

Observe the original formula and find [C ^ B + (a-c) ^ B] % A = 0.

(Proof: Just split (a-c) ^ B with the binary theorem)

So it is easy: When a is an odd number, the original formula is 0; When a is an even number, the original formula is = (a/2) ^ B %. and then quickly modulo out the solution.

Click it by yourself:

#include <cstdio>using namespace std;typedef long long ll;int mod;ll mypow(ll a, ll n){    ll ret = 1;    while(n)    {        if(n%2) ret = (ret*a)%mod;        a = (a*a)%mod;        n >>= 1;    }    return ret;}int cal(int a, int b){    if(a&1) return 0;    return mypow(a/2,b);}int main(){    int a,b;    while(scanf("%d %d",&a,&b)==2)    {        mod = a;        printf("%d\n",cal(a,b));    }}

It's still a pattern...