1745:Divisibility
- 总时间限制:
- 1000ms
- 内存限制:
- 65536kB
- 描述
- 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. - 输入
- 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. - 输出
- Write to the output file the word "Divisible" if given sequence of integers is divisible by K or "Not divisible" if it's not.
- 样例输入
-
4 7 17 5 -21 15
- 样例输出
-
Divisible
- 来源
Northeastern Europe 1999
代码:
#include<cstdio>
#include<cstring>
int a[100],b[100];
int main(){
int n,k,tmp;
scanf("%d%d",&n,&k);
scanf("%d",&tmp);
a[(tmp%k+k)%k] = 1;
for(int i = 2; i <= n; ++i){
memset(b,0,sizeof(b));
scanf("%d",&tmp);
for(int j = 0; j < k; j++){
if(a[j]){
b[((j+tmp)%k+k)%k] = 1;
b[((j-tmp)%k+k)%k] = 1;
}
}
memcpy(a,b,sizeof(int)*k);
}
if(a[0]) puts("Divisible");
else puts("Not divisible");
return 0;
}