思路：设分母为：dwn,x^p的个数为mp[p]。分子为 x^(a1 + a2 …… + an - ai)相加。因为an最大，所以分子第n项最小。提出该项dwn / (x^an) == x((分母次数) - an)，设总次数为（分母次数 - div）如果该项系数（个数）% x == 0,则mp[div - 1+=mp[div] / x;div--；
#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
#define N 500010
#define maxn (N + 10)
#define Max(a,b) (a > b ? a : b)
#define Min(a,b) (a < b ? a : b)
typedef long long ll;
const ll INF = (1e9) + 10;
const ll mod = (1e9) + 7;
unordered_map <int,int> mp;
ll f_pow(ll x,ll p){
ll ans = 1;
while(p){
if(p % 2 == 1){
ans *= x;ans %= mod;
}
x *= x;x %= mod;p /= 2;
}
return ans;
}
int main() {
int n, x;scanf("%d%d",&n, &x);
int an = 0;
ll sum = 0;
for(int i = 1;i <= n;i++){
int a;scanf("%d",&a);
mp[a]++;sum += a;
if(i == n)an = a;
}
int div = an;
while(mp[div] % x == 0 && div >= 0){
mp[div - 1] += (mp[div] / x);div--;
}
if(div < 0)div = 0;
sum -= div;
//printf("sum = %lld\n",sum);
ll ans = f_pow(x,sum);
printf("%lld\n",ans);
return 0;
}
原文：https://www.cnblogs.com/cleanerhgf/p/11991705.html