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POJ1745Divisibility【dp】

2015-08-26 10:28 246 查看

Divisibility

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11053		Accepted: 3953

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

题意：给出N个数通过在其中添加+-号问是否能计算出k的倍数

dp[i][j]代表前i个数余数为j是否存在

#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
const int MAX=10010;
int dp[MAX][110];
int num[MAX];
int main()
{
int n,m,k,i,j;
while(scanf("%d%d",&n,&k)!=EOF){
memset(dp,0,sizeof(dp));
for(i=0;i<n;++i){
scanf("%d",&num[i]);
if(num[i]<0){
num[i]*=-1;
}
num[i]%=k;
}
dp[0][num[0]]=1;
for(i=1;i<n;++i){
for(j=0;j<k;++j){
if(dp[i-1][j]){
dp[i][(j+num[i])%k]=1;
dp[i][(k+j-num[i])%k]=1;
}
}
}
if(dp[n-1][0])
printf("Divisible\n");
else
printf("Not divisible\n");
}
return 0;
}

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标签： POJ1745Divisibilityd

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