DP PKU 1754

开个二维数组。第一维保存几个数。第二维保存的是余数，若dp【】【】=1，就代表存在

负数取模的时候要注意。

代码虐我千百遍，我待代码如初恋


Divisibility

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10355		Accepted: 3685

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

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <ctype.h>
#include <limits.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <set>
#include <map>
using namespace std;
int dp[10010][110];//前面的代表有几个数，后面的代表前几个数取余之后的结果
int num[10010];

int main(){
int n,k;
int i,j;

while(~scanf("%d%d",&n,&k)){
memset(dp,0,sizeof(dp));
memset(num,0,sizeof(num));
for(i=1;i<=n;i++){
scanf("%d",&num[i]);
}
dp[0][0] = 1;
for(i=1;i<=n;i++){
for(j=0;j<=k-1;j++){
if(dp[i-1][j] == 1){
int t = ((j+num[i])%k+k)%k;
dp[i][t] = 1;
int t1 = ((j-num[i])%k+k)%k;
dp[i][t1] = 1;
}
}
}
if(dp
[0] == 1){
printf("Divisible\n");
}
else{
printf("Not divisible\n");
}
}

return 0;
}