C. Prime Number----数论+快速幂

C. Prime Number time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Simon has a prime number x and an array of non-negative integers a1, a2, ..., an.

Simon loves fractions very much. Today he wrote out number  on a piece of paper. After Simon led all fractions to a common denominator and summed them up, he got a fraction: , where number t equals xa1 + a2 + ... + an. Now Simon wants to reduce the resulting fraction.

Help him, find the greatest common divisor of numbers s and t. As GCD can be rather large, print it as a remainder after dividing it by number 1000000007 (109 + 7).

Input

The first line contains two positive integers n and x (1 ≤ n ≤ 105, 2 ≤ x ≤ 109) — the size of the array and the prime number.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ a1 ≤ a2 ≤ ... ≤ an ≤ 109).

Output

Print a single number — the answer to the problem modulo 1000000007 (109 + 7).

Examples input 2 2 2 2 output 8 input 3 3 1 2 3 output 27 input 2 2 29 29 output 73741817 input 4 5 0 0 0 0 output 1 Note

In the first sample . Thus, the answer to the problem is 8.

In the second sample, . The answer to the problem is 27, as 351 = 13·27, 729 = 27·27.

In the third sample the answer to the problem is 1073741824 mod 1000000007 = 73741817.

In the fourth sample . Thus, the answer to the problem is 1.

题目链接：http://codeforces.com/contest/359/problem/C

想了40分钟的我，还是去看了博客，所有思路都不通。。。

http://blog.csdn.net/blesslzh0108/article/details/61920030

代码：

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define inf 1000000007 #define LL long long using namespace std; LL a[100005]; LL power(LL x,LL m){ LL ans=1; x=x%inf; while(m){ if(m&1) ans=(ans*x)%inf; x=(x*x)%inf; m/=2; } return ans%inf; } int main(){ LL n,x; LL s=0; scanf("%I64d%I64d",&n,&x); for(int i=0;i<n;i++){ scanf("%I64d",&a[i]); s+=a[i]; } for(int i=0;i<n;i++){ a[i]=s-a[i];//求出分子的幂数 } a[n]=-1; sort(a,a+n); LL cnt=1,ans; for(int i=1;i<=n;i++){ if(a[i]!=a[i-1]){ if(cnt%x==0){//如果系数可以整除x，则在前者+1，前者值发生改变，i--再判断一次 cnt/=x; a[i-1]+=1; i--; } else{//不相等，且系数不能整除x，即为所求 ans=a[i-1]; break; } } else{//如果相等，记录加一 cnt++; } } ans=min(ans,s); printf("%I64d\n",power(x,ans)); }