E - Fantasy of a Summation LightOJ1213-白红宇

E - Fantasy of a Summation LightOJ1213

阅读量：5462 次

发布时间：2019-06-16

本文共 2686 字，大约阅读时间需要 8 分钟。

E - Fantasy of a Summation

Time Limit:

2000/1000 MS (Java/Others) Memory Limit: 128000/64000 KB (Java/Others)

Submit

Problem Description

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.

Actually the code was about: 'You are given three integers n, K, MOD and n integers: A0, A1, A2 ... An-1, you have to write K nested loops and calculate the summation of all Ai where i 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 contains n non-negative integers denoting A0, A1, A2 ... An-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

23 1 350001 2 32 3 350001 2

Sample Output

Case 1: 6Case 2: 36

1 #include 
        
          2 #include 
         
           3 #include 
          
            4 #include 
           
             5 #include 
            
              6 #include 
             
               7 #include 
               8 #include 
               
                 9 #include 
                
                 10 #include 
                 
                  11 #include 
                  
                   12 using namespace std;13 typedef long long LL;14 const int N=1e5+10;15 const int INF=0x3f3f3f3f;16 int cas=1,T;17 int n,k,mod;18 int quick_mod(int t,int n)19 {20 int ans=1;21 while(n)22 {23 if(n&1) ans=ans*t%mod;24 t=t*t%mod;25 n>>=1;26 }27 return ans;28 }29 int main()30 {31 //freopen("1.in","w",stdout);32 // freopen("1.in","r",stdin);33 // freopen("1.out","w",stdout);34 scanf("%d",&T);35 while(T--)36 {37 scanf("%d%d%d",&n,&k,&mod);38 int ans=0;39 for(int i=0,x;i

solve.cpp

题解：

仔细研究一下代码会发现只是一条公式：k*n^(k-1)*(a1+a2+...+an)

然后快速幂即可，注意不要被循环绕晕

转载于:https://www.cnblogs.com/cdyboke/p/7010529.html

你可能感兴趣的文章

火狐无法访问本机IIS部署的网站，弹出：此地址使用了一个通常用于网络浏览以外目的的端口.出于安全原因,Firefox 取消了该请求的解决办法...