题解：根据题目给的程序，就是计算给的这个序列，进行k次到n的循环，每个数需要加的次数是k*n^(k-1)，所以快速幂取模，算计一下就可以了。
代码语言：javascript复制#include <bits/stdc .h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f3f;
long long pow_mod(ll a, ll k, ll mod)
{
ll ans = 1;
while(k)
{
if(k%2)
ans *= a;
ans %= mod;
a = a * a;
a %= mod;
k /=2;
}
return ans;
}
int main()
{
int T;
ll n,k,mod,x,sum;
while(~scanf("%d",&T))
{
int cas = 1;
while(T--)
{
sum = 0;
scanf("%lld%lld%lld",&n,&k,&mod);
for(ll i = 0; i < n; i )
{
scanf("%lld",&x);
sum = (x * (k * pow_mod(n,k-1,mod)%mod)%mod);
sum %= mod;
}
printf("Case %d: %lldn",cas , sum);
}
}
return 0;
}Problem: 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: %dn", caseno, res); } return 0; } 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
