ACdream 1007

input

T           <=10

n k            n<=1000         k<=10^18

a1,a2,...an                |ai|<=10^18

output

(a1^k+a2^k+...+an^k)%10^10+7

Sample Input

2

3 1

1 2 3

3 10

1 2 3

Sample Output

6 60074

做法：快速幂+__int128,需注意ai可能是负数，模的是10^10+9，超int了

模乘法：对一个数可以拆分为N=（P*(N/P)+(N%P)）

#include <bits/stdc++.h>
#define MAX 100000
#define LL long long
//#define __int128 long long
#define mod 10000000007LL
using namespace std;
int cas=1,T;
template <class T>  
bool scanff(T &ret)
{ //Faster Input  
    char c; int sgn; T bit=0.1;  
    if(c=getchar(),c==EOF) return 0;  
    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();  
    sgn=(c=='-')?-1:1;  
    ret=(c=='-')?0:(c-'0');  
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');  
    if(c==' '||c=='
'){ ret*=sgn; return 1; }  
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;  
    ret*=sgn;  
    return 1;  
} 
template <class T>
void printff(T ret)
{ //Faster Output
    char s[22];
    if(ret<0) { printf("-");ret=-ret; }
    int len=0;
    while(ret)
    {
        s[++len]=ret%10+'0';
        ret/=10;
    }
    if(!len) s[++len]='0';
    while(len)
    {
        printf("%c",s[len]);
        len--;
    }
}
LL qmod( __int128 a, LL n)
{
    __int128 res=1;
    LL flag=1;
    if(a<0) { flag=(n&1?-1:1);a=-a; }
    a%=mod;
    while(n)
    {
        if(n&1) res=res*a%mod;
        a=a*a%mod;
        n>>=1;
    }
    return (LL)res*flag;
}
int main()
{
    //freopen("1.in","w",stdout);
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    scanf("%d",&T);
    while(T--)
    {
        int n;
        LL k,a,res=0;
        scanf("%d%lld",&n,&k);
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&a);
            res=(res+qmod(a,k))%mod;
        }
        printff((res+mod)%mod);
        printf("
");
    }
    //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
    return 0;
}