Prime Number
CodeForces - 359CSimon 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).
OutputPrint a single number — the answer to the problem modulo 1000000007 (109 + 7).
Examples2 2
2 2
8
3 3
1 2 3
27
2 2
29 29
73741817
4 5
0 0 0 0
1
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.
OJ-ID:
CodeForce 359C
author:
Caution_X
date of submission:
20191004
tags:
数论
description modelling:
计算s,t的gcd。
输入a0,a1,a2,.....an,x
s=x^(a0+a1+a2+....+an).
t=x^(a1+a2+...+an)+x^(a0+a2+a3+....+an)+.....+x^(a0+a1+a2+.....+a(n-1)).
major steps to solve it:
1.提取分子上幂最小的数(根据t,s的关系可知该数一定可以和s整除)
2.比较提取的幂和(a0+a1+...an)的大小
3.快速幂取模输出答案
warnings:
分子可能比分母大
AC Code:
#include<algorithm> #include<cstdio> #include<iostream> #include<map> using namespace std; typedef long long ll; ll mod=1e9+7; map<ll,ll> book; map<ll,ll>::iterator it; ll a[100005]; ll qp(ll a,ll b) { ll ans=1; while(b) { if(b&1) { ans*=a; ans%=mod; } a*=a; a%=mod; b>>=1; } return ans; } int main() { ll n,x; scanf("%lld%lld",&n,&x); ll sum=0; for(int i=0;i<n;i++) { scanf("%lld",&a[i]); sum+=a[i]; } for(int i=0;i<n;i++) { book[sum-a[i]]++; } for(it=book.begin();it!=book.end();it++) { if(it->second>=x) { ll tmp=it->second/x; book[it->first+1]+=tmp; book[it->first]-=tmp*x; } } it=book.begin(); while(it->second==0) it++; printf("%lld\n",qp(x,min(it->first,sum))); return 0; }