C++(clang++11) 解法, 执行用时: 680ms, 内存消耗: 1412K, 提交时间: 2020-11-01 10:52:43

#include<iostream>
using namespace std;
const long long mod=1e9+7;
#define ll long long
ll a[200000];
int main()
{
	ll n,x;
	while(cin>>n>>x) {
		for(ll i=1;i<=n;i++) {
			cin>>a[i];
		}
		ll sum=x,k;
		for(ll i=1;i<=n;i++) {
			cin>>k;
			sum=(sum+k*a[i])%mod; 
		}
		for(ll i=1;i<n;i++) {
			sum=(sum*x)%mod;
		}
		cout<<sum<<endl;
	}
	return 0;
}

Python3(3.9) 解法, 执行用时: 1040ms, 内存消耗: 22340K, 提交时间: 2020-11-11 21:54:15

def gt(): return list(map(int,input().split()))
M=1000000007
while 1:
    try:
        n,x = gt()
    except:
        break;
    a = gt()
    b = gt()
    ans = pow(x,n-1,M)
    tmp = x
    for i in range(n):
        tmp += a[i]*b[i]
        tmp %= M
    ans = ans*tmp%M
    print(ans)

pypy3(pypy3.6.1) 解法, 执行用时: 1073ms, 内存消耗: 62280K, 提交时间: 2020-12-30 20:35:58

MODN=int(1e9+7)
while True:
	try:
		n,x=list(map(int,input().split()))
	except:
		break
	a=list(map(int,input().split()))
	b=list(map(int,input().split()))
	up=0
	for i in range(n):
		up+=a[i]*b[i]
	up%=MODN
	print((pow(x,n,MODN)+pow(x,n-1,MODN)*up)%MODN)