POJ Divisibility 1745【动态规划】

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Divisibility

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 11045 Accepted: 3950

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

4 717 5 -21 15

Sample Output

Divisible

Source

Northeastern Europe 1999

题意：给你n个数，经过加减的操作，看能不能整除K。

解题思路：

简单dp：dp[ i ][ j ]  代表前 i 个数字整除 m 的余数为 j。

AC代码：

#include <stdio.h>#include <math.h>#include <vector>#include <queue>#include <string>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>using namespace std;int dp[10010][100];int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF){        int a[10010];        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        dp[0][0]=1;        for(int i=1;i<=n;i++){            for(int j=0;j<m;j++){                if(dp[i-1][j]){                    dp[i][((j-a[i])%m+m)%m]=1;                    dp[i][((j+a[i])%m+m)%m]=1;                }            }        }        if(dp[n][0]) printf("Divisible\n");        else printf("Not divisible\n");    }    return 0;}