HDU4506+快速求幂

solve(a,b,c)=a^b%c

/*
快速求幂
solve（a,b,c）==a^b%c
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<math.h>
using namespace std;
typedef __int64 int64;
const int maxn = 10005;
const int64 mod = 1000000007;
int64 a[ maxn ],b[ maxn ];
int solve( int64 m,int64 n,int64 mod ){
    int sum = 1;
    while( n>=1 ){
        if( n%2==1 ){
            sum = sum*m%mod;
        }
        m = m*m%mod;
        n/=2;
    }
    return sum%mod;
}    
int main(){
    int ca;
    scanf("%d",&ca);
    while( ca-- ){
        int n;
        int64 t,k;
        scanf("%d%I64d%I64d",&n,&t,&k);
        for( int i=0;i<n;i++ )
            scanf("%I64d",&a[ i ]);
        int64 t1,t2;
        t1=t/n;
        t2=t%n;
        int64 tt = solve( k,t,mod );
        for( int i=0;i<n;i++ ){
            if( t2<= i ) b[ i ] = a[ i-t2 ]*tt%mod;
            else b[ i ] = a[ n-( t2-i ) ]*tt%mod;
        }
        for( int i=0;i<n;i++ )
            if( i==0 ) printf("%d",b[ i ]);
            else printf(" %d",b[ i ]);
        printf("\n");
    }
    return 0;
}