modulo 求余 determine 确定 determinant n行列式
the input consists of several test cases terminated by end-of-file
terminal n终端 terminate v终结 the sum of ... 总和
输出描述:
For each test case, print an integer which denotes the result.
示例1
输入
复制
2 1 0 0 0 0 2 1 1000000000 1000000000 1000000000 1000000000 3 2 2 3 3 2 3 3
输出
复制
1 99 96
#include<iostream>
using namespace std;
#define ll long long
#define li __int128
const int mod=1e9+7;
const int maxn=1e6+10;
ll n,x,a[maxn],b;int main(){while(cin>>n>>x){for(int i=1;i<=n;i++){scanf("%lld",&a[i]);}ll ans=0;for(int i=1;i<=n;i++){scanf("%lld",&b);a[i]=(b*a[i])%mod;ans=(ans+a[i])%mod;}ans=(ans+x)%mod;for(int i=1;i<n;i++){ans=(ans*x)%mod;}printf("%lld\n",ans%mod);} return 0;
}