E - Fantasy of a Summation LightOJ1213

Posted 2020-09-22 dy blog

E - Fantasy of a Summation

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

Submit Status

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.

 1 #include <stdio.h>
 2  
 3 int cases, caseno;
 4 int n, K, MOD;
 5 int A[1001];
 6  
 7 int main() {
 8     scanf("%d", &cases);
 9     while( cases-- ) {
10         scanf("%d %d %d", &n, &K, &MOD);
11  
12         int i, i1, i2, i3, ... , iK;
13  
14         for( i = 0; i < n; i++ ) scanf("%d", &A[i]);
15  
16         int res = 0;
17         for( i1 = 0; i1 < n; i1++ ) {
18             for( i2 = 0; i2 < n; i2++ ) {
19                 for( i3 = 0; i3 < n; i3++ ) {
20                     ...
21                     for( iK = 0; iK < n; iK++ ) {
22                         res = ( res + A[i1] + A[i2] + ... + A[iK] ) % MOD;
23                     }
24                     ...
25                 }
26             }
27         }
28         printf("Case %d: %d\n", ++caseno, res);
29     }
30     return 0;
31 }

 

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

2
3 1 35000
1 2 3
2 3 35000
1 2

Sample Output

Case 1: 6
Case 2: 36

 1 #include <stdio.h>
 2 #include <queue>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <map>
 8 #include <set>
 9 #include <ctime>
10 #include <cmath>
11 #include <cctype>
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<n;i++)
40         {
41             scanf("%d",&x);
42             ans+=x%mod;
43         }
44         ans%=mod;
45         ans=k%mod * quick_mod(n,k-1) % mod *ans %mod;
46         printf("Case %d: %d\n",cas++,ans);
47     }
48     //printf("time=%.3lf\n",(double)clock()/CLOCKS_PER_SEC);
49     return 0;
50 }

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