POJ 1745 Divisibility （DP）

Divisibility

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 11452 Accepted: 4097

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

真心想不到这题也能用DP做，DP太强大了。题意：给你一列整数，在整数间加‘ + ’ 或 ‘ - ‘，使这个算式的值能被k整除。思路：可以把结果mod k看作状态，这样虽然n个数有2^n-1种运算方式，但结果只有k种，所以只需枚举这k个数即可。dp[i][j]表示 前i个数运算结果mod k是否存在余数j，转移方程就简单了：如果dp( i-1, j)为true，那么把dp(i, (j+a[i])mod k)和

dp(i, (j-a[i])mod k)置真。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int dp[10010][110],num[10010];int main(){int n,k,i,j;while(scanf("%d%d",&n,&k)!=EOF){for(i=1;i<=n;i++)scanf("%d",&num[i]);memset(dp,false,sizeof(dp));dp[0][0]=true;for(i=1;i<=n;i++){for(j=0;j<k;j++){if(dp[i-1][j]){dp[i][abs(j+num[i])%k]=true;dp[i][abs(j-num[i])%k]=true;}} } if(dp[n][0]) printf("Divisible\n"); else printf("Not divisible\n"); }return 0;}