codeforces 1362 E 进制的性质

写法想多，对结果无优化的方法不必要多写。

 

最开始思路是想把k数组的偶数个相同的消除，

6 2

1  2  3  3 3 4

即变成1 2 3 4，然后从大到小排序，因为

 

 然后用最大的那个数pow减去后面的即可。

但是徒增操作量，因为倘若n个数都不相同，则时间复杂度不变。

所以可以直接从大到小排序，然后直接计算即可。

还有一个细节

 

 

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '\n'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll mod = 1e9+7;
ll MOD = 1e9+3;
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 1e6+10;
bool cmp(ll a,ll b)
{
    return a > b;
}
ll k[maxn];
int main()
{
    //freopen("input.txt", "r", stdin);
    int _;
    scanf("%d",&_);
    while(_--)
    {
        ll n,p;
        cin>>n>>p;
        rep(i,1,n) cin>>k[i];
        if(p==1){printf("%d\n",n%2);continue;}
        sort(k+1,k+1+n,cmp);    
        ll ans = 0,res = 0;
        rep(i,1,n)    {
            if(ans == 0 && res == 0) {
                ans = powmod(p,k[i],mod);
                res = powmod(p,k[i],MOD);
            }
            else {
                ans = (ans - powmod(p,k[i],mod) + mod)%mod;
                res = (res - powmod(p,k[i],MOD) + MOD)%MOD;
            }
        }
        cout<<ans<<endl;
    }
}