Codeforces Round #209 (Div. 2) C - Prime Number

传送门

题意

给出n个数及x，求

\[\frac{\sum _{i=1}^n x^{a_1+a_2+...+a_{i-1}+a_{i+1}+...a_n}}{\prod_{i=1}^n x^{a_i}}
\]

分析

结果必然为\(x^{sum}\),sum的值首先取所有数的和减去最大值

然后暴力合并，具体原因我不太懂，只能附上CF的标准题解

Obviously, the answer is \(x^v\). Let \(sum = a1 + a2 + ... + an\). Also let $si = sum - ai $(the array of degrees). After that let's find value \(v\) by the following algorithm: Let's consider a sequence of degrees as decreasing sequence. Now we will perform the following operation until it's possible to perfom it. Take the minimum degree \(v\) from the array of degrees and calculate the number of elements \(cnt\), which have the same degree. If \(cnt\) multiples of \(x\), then replace all \(cnt\) elements by \(cnt / x\) elements of the form \(v + 1\). Since the sequence of degrees is a decreasing sequence, we can simply assign them to the end. If \(cnt\) is not a multiple of \(x\), then we found the required value \(v\). Also you need to check, that \(v\) is not greater then sum. Otherwise, \(v\) will be equals to sum.

详情见代码

代码

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <string>
#include <map>
#include <queue>
using namespace std;

#define ll long long
#define F(i,a,b) for(int i=a;i<=b;++i)
#define R(i,a,b) for(int i=a;i<b;++i)
#define mem(a,b) memset(a,b,sizeof(a))
#pragma comment(linker, "/STACK:102400000,102400000")
inline void read(int &x){x=0; char ch=getchar();while(ch<'0') ch=getchar();while(ch>='0'){x=x*10+ch-48; ch=getchar();}}
const ll mod=1e9+7;
int n,x,a[100100],num,t;
ll sum,cnt;
bool vis[100100];
ll mi(int x,ll cnt)
{
    ll ret=x,ans=1;
    while(cnt)
    {
        if(cnt&1) ans=(ans*ret)%mod;
        cnt>>=1,(ret*=ret)%=mod;
    }
    return ans;
}
int main()
{
    scanf("%d %d",&n,&x);
    F(i,1,n){ scanf("%d",a+i);sum+=(ll)a[i]; }
    sum-=a[n];t=a[n];
    num=0;
    F(i,1,n)
    {
        if(a[i]==a[n]) { num++;vis[i]=1; }
        a[i]=a[n]-a[i];
    }
    while(num%x==0&&t)
    {
            sum++;num/=x;t--;
            F(i,1,n) if(!vis[i])
            {
                a[i]--;
                if(a[i]==0) {num++;vis[i]=1;}
            }
        if(num==0) break;
    }
    //printf("%d\n",sum);
    printf("%lld\n",mi(x,sum));
    return 0;
}