codeforces 359 C (数论)

题意：给定质数x,有n个1/(x^ai) (1<=i<=n)相加，分母为x^(a1 + a2 + a3 + …… + an),分子为T。求分子，分母最大公因数。

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;
}