Fantasy of a Summation

Fantasy of a Summation

If you think codes, eat codes then sometimes you may get stressed. In your dreams you may see huge codes, as I have seen once. Here is the code I saw in my dream.

#include <stdio.h>

int cases, caseno;
int n, K, MOD;
int A[1001];

int main() {
    scanf("%d", &cases);
    while( cases-- ) {
        scanf("%d %d %d", &n, &K, &MOD);

        int i, i1, i2, i3, ... , iK;

        for( i = 0; i < n; i++ ) scanf("%d", &A[i]);

        int res = 0;
        for( i1 = 0; i1 < n; i1++ ) {
            for( i2 = 0; i2 < n; i2++ ) {
                for( i3 = 0; i3 < n; i3++ ) {
                    ...
                    for( iK = 0; iK < n; iK++ ) {
                        res = ( res + A[i1] + A[i2] + ... + A[iK] ) % MOD;
                    }
                    ...
                }
            }
        }
        printf("Case %d: %d\n", ++caseno, res);
    }
    return 0;
}

Actually the code was about: 'You are given three integers n,K, MOD and n integers: A₀, A₁, A₂ ... A_n-1, you have to writeK nested loops and calculate the summation of all A_i wherei is the value of any nested loop variable.'

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with three integers: n (1 ≤ n ≤ 1000), K (1 ≤ K < 2³¹), MOD (1 ≤ MOD ≤ 35000). The next line containsn non-negative integers denoting A₀, A₁, A₂ ... A_n-1. Each of these integers will be fit into a 32 bit signed integer.

Output

For each case, print the case number and result of the code.

Sample Input

2

3 1 35000

1 2 3

2 3 35000

1 2

Sample Output

Case 1: 6

Case 2: 36

这种题可能算一种类型吧，即推公式，算出每个值对于 答案   的贡献   ，或者说是每个值被选中的概率 。

题目很长，就是让你求给定的那段代码所能求出的值，显然你不能用他写的那段代码。

从给定的代码上可以看出： 每次进行一个 n  的循环，即进行了 n^k  次的加法，而每次的求和都是重复的选取n个数，所以每个数的贡献就是k*n^(k-1);

所以实现就是用个快速幂遍历一遍。

代码：




#include<iostream>#include<cstdio>#include<cmath>using namespace std;typedef long long ll;ll n,k,mod;ll Pow(ll a,ll b){    ll ans=1;    a%=mod;    while(b)    {        if(b&1)            ans=ans*a%mod;        b>>=1;        a=a*a%mod;    }    return ans;}int main(){    int T;    int kase=1;    scanf("%d",&T);    while(T--)    {        scanf("%lld%lld%lld",&n,&k,&mod);        ll xx=0,x;        for(ll i=0; i<n; i++)        {            scanf("%lld",&x);            xx+=x;        }        xx%=mod;        ll sum=(k*Pow(n,(k-1))*xx)%mod;        printf("Case %d: ",kase++);        printf("%lld\n",sum);    }    return 0;}