UVA 10036 Divisibility

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本文探讨了一组整数通过加减运算形成的数学表达式，并着重研究了如何判断这些表达式的结果是否能被某个整数整除。通过实例演示了计算过程，并提供了一个基于动态规划的方法来解决此类问题。

Consider an arbitrary sequence of integers. One can place + or - operators between integers in the sequence, thus deriving different arithmetical expressions that evaluate to different values. Let us, for example, take the sequence: 17, 5, -21, 15. There are eight possible expressions:

17	+	5	+	-21	+	15	=	16
17	+	5	+	-21	-	15	=	-14
17	+	5	-	-21	+	15	=	58
17	+	5	-	-21	-	15	=	28
17	-	5	+	-21	+	15	=	6
17	-	5	+	-21	-	15	=	-24
17	-	5	-	-21	+	15	=	48
17	-	5	-	-21	-	15	=	18

We call the sequence of integers divisible by K if + or - operators can be placed between integers in the sequence in such way that resulting value is divisible by K. In the above example, the sequence is divisible by 7 (17+5+-21-15=-14) but is not divisible by 5.

You are to write a program that will determine divisibility of sequence of integers.

Input

The first line of the input file contains a integer M indicating the number of cases to be analyzed. Then M couples of lines follow.
For each one of this couples, the first line contains two integers, N and K (1 <= N <= 10000, 2 <= K <= 100) separated by a space. The second line contains a sequence of N integers separated by spaces. Each integer is not greater than 10000 by it's absolute value.

Output

For each case in the input file, write to the output file the word "Divisible" if given sequence of integers is divisible by K or "Not divisible" if it's not.

Sample input

2
4 7
17 5 -21 15
4 5
17 5 -21 15

Sample Output

Divisible
Not divisible

dp[i][j]中的i表示加上或者减去第i个数时，余数为j是否为true。最后得到dp[n-1][0]是否为真。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

int dp[11000][110];

int main(){
//freopen("test.in","rt",stdin);
    int t,n,k,a,i,j;
    cin >>t;
    while (t--){
        cin >>n >>k >>a;
        memset(dp, 0, sizeof(dp));
        dp[0][(a % k + k) % k] = 1;
        for (i = 1; i < n; ++i){
            cin >>a;
        a = abs(a) % k;
        for (j = 0; j < k; ++j)
            if (dp[i - 1][j])
                dp[i][(j + a) % k] = dp[i][(j - a + k) % k] = 1;
        }
        cout <<(dp[n - 1][0] ? "Divisible" : "Not divisible") <<endl;
    }
    return 0;
}