POJ 1745 Divisibility 動態規劃

 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

用到一個公式

（a + b）  % c = (a % c + b % c) % c

當a小於0時  a對c的餘數 是 a % c + c

一個簡單的DP，不用把數字存起來，直接算就好。

AC代碼 ： 

使用了位運算來簡化空間複雜度，也可以使用滾動數組。（位運算的原理類似於滾動數組，把int的前兩位當作第一層數據和第二層然後每次結束右移單位1，覆蓋數據）

#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;  //無論t的正負取其餘數
        dp[t] = 1;
        fn(i,n - 1){
            cin >> t;
            t %= k;
            if(t == 0) continue; //其實t = 0的話也可以接着算，不過爲了省時間還是判斷一下
            fn(j,k) if(dp[j] & 1){    //判斷上一個數時是否有餘數
                    dp[(j + t + k) % k] |= 2; //第二位，置1
                    dp[(j - t + k) % k] |= 2;
                }
            fn(j,k) dp[j] >>= 1;  //覆蓋上一組數據
        }
        if(dp[0])cout << "Divisible" << endl;
        else cout << "Not divisible" << endl;
    }
    return 0;
}