Fantasy of a Summation

TimeLimit: 2000ms  MemoryLimit:32768KB
64-bit integer IO format:%lld
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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.

#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 ... 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.

SampleInput
2
3 1 35000
1 2 3
2 3 35000
1 2
SampleOutput
Case 1: 6
Case 2: 36
Submit
题目统计信息详细
总AC数5
通过人数4
尝试人数5
总提交量20
AC率20.00%
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