lightoj1213

maksyuki 发表于 oj 分类，标签:
27 8月 2015
0

1213 - Fantasy of a Summation

#include <stdio.h>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.

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 ... 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

Output for Sample Input
23 1 350001 2 32 3 35000

1 2

 Case 1: 6Case 2: 36

 

题目类型：简单数学 

算法分析：可以直接计算出每一位中可能出现的数的和，然后再乘以位数k就可以了，公式为sum * (n ^ (k - 1)) * k % mod，sum表示一位中所有数的总和，n ^ (k - 1)表示在一位中出现一种数字的次数，由于数据比较大，所以需要使用整数快速幂取模来计算