POJ 1745 Divisibility 动态规划

原創

胖胖好酷

2020-02-21 09:56

Divisibility

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1745

Description

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 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

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

4 7
17 5 -21 15

Sample Output

Divisible

用到一个公式

（a + b） % c = (a % c + b % c) % c

当a小于0时 a对c的余数是 a % c + c

一个简单的DP，不用把数字存起来，直接算就好。

AC代码：

使用了位运算来简化空间复杂度，也可以使用滚动数组。（位运算的原理类似于滚动数组，把int的前两位当作第一层数据和第二层然后每次结束右移单位1，覆盖数据）

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

#define fn(i,n) for(int i = 0; i < n; i++)
int dp[110];

int main(){
    int n, k, t;
    while(scanf("%d %d",& n,& k) != EOF){
        memset(dp,0,sizeof(dp));
        cin >> t;
        t = (t % k + k) %k;  //无论t的正负取其余数
        dp[t] = 1;
        fn(i,n - 1){
            cin >> t;
            t %= k;
            if(t == 0) continue; //其实t = 0的话也可以接着算，不过为了省时间还是判断一下
            fn(j,k) if(dp[j] & 1){    //判断上一个数时是否有余数
                    dp[(j + t + k) % k] |= 2; //第二位，置1
                    dp[(j - t + k) % k] |= 2;
                }
            fn(j,k) dp[j] >>= 1;  //覆盖上一组数据
        }
        if(dp[0])cout << "Divisible" << endl;
        else cout << "Not divisible" << endl;
    }
    return 0;
}