LightOJ1213 Fantasy of a Summation —— 快速幂

题目链接：https://vjudge.net/problem/LightOJ-1213

1213 - Fantasy of a Summation

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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 write K nested loops and calculate the summation of all A_i 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 ... 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	Output for Sample Input
2 3 1 35000 1 2 3 2 3 35000 1 2	Case 1: 6 Case 2: 36

 

题解：

根据代码， 可知每个数在特定位置中出现了 n^(k-1)次，而总共有k个位置，所以每个数出现了k*n^(k-1)次，所以答案为： ∑ a[i]*k*n^(k-1)，1<=i<=n。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 //const int MOD = 35000+7;
 const int MAXN = 1e6+;

 int MOD;
 LL qpow(LL x, LL y)
 {
     LL s = ;
     while(y)
     {
         if(y&) s = (s*x)%MOD;
         x = (x*x)%MOD;
         y >>= ;
     }
     return s;
 }

 int main()
 {
     int T, n, k, kase = ;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%d%d%d", &n,&k,&MOD);
         LL sum = ;
         for(int i = ; i<=n; i++)
         {
             LL val;
             scanf("%lld", &val);
             sum = (sum+(val%MOD))%MOD;
         }

         LL ans = (((1LL*sum*k)%MOD)*qpow(n, k-))%MOD;
         printf("Case %d: %lld\n", ++kase, ans);
     }
 }