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
There are multiple test cases, the first line is the number of test cases.
The first line of each test case 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
2
4 7
17 5 -21 15
4 5
17 5 -21 15
Sample Output
Divisible
Not divisible
啟發(fā)：1，涉及整除就要聯(lián)想取模！！！！！！！！
2，這是有層次性，不要搞混，以下是錯(cuò)誤代碼
#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
int testnum;
int num,pos;
cin>>testnum;
int a[101],temp,i,j;
while(testnum--)
{
memset(a,0,sizeof(a));
cin>>num>>pos;
cin>>temp;
temp%=pos;
if(temp<0)temp+=pos;
a[temp]=1;
for(i=1;i<num;i++)
{
cin>>temp;
temp%=pos;
if(temp<0)temp+=pos;//temp%pos¿ÉÄÜÊǸºÊý
for(j=0;j<pos;j++)
{
if(a[j]>0)
{
a[(j+temp)%pos]++;
if((j-temp)<0)
{
a[(j-temp)+pos]++;
}
else
{
a[j-temp]++;
}
if(temp!=0)
a[j]=0;
}
}
}
if(a[0]>0)cout<<"Divisible"<<endl;
else cout<<"Not divisible"<<endl;
}
//system("PAUSE");
return 0;
}
這是有層次性，再添加一個(gè)元素時(shí)，只能改變前一組a[j]的值
if(a[j]>0)//
{
a[(j+temp)%pos]++;//這里的修改應(yīng)該不應(yīng)添加到前面的a[j]中去。
if((j-temp)<0)
所以應(yīng)當(dāng)用兩個(gè)數(shù)組。
#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
int testnum;
int num,pos;
cin>>testnum;
int a[101],temp,i,j;
int b[101];
while(testnum--)
{
memset(a,0,sizeof(a));
memset(a,0,sizeof(b));
cin>>num>>pos;
cin>>temp;
temp%=pos;
if(temp<0)temp+=pos;
a[temp]=1;
for(i=1;i<num;i++)
{
cin>>temp;
temp%=pos;
if(temp<0)temp+=pos;//temp%pos¿ÉÄÜÊǸºÊý
for(j=0;j<pos;j++)
{
if(a[j]>0)
{
b[(j+temp)%pos]++;
if((j-temp)<0)
{
b[(j-temp)+pos]++;
}
else
{
b[j-temp]++;
}
}
}
memset(a,0,sizeof(a));
for(j=0;j<pos;j++)
{
if(b[j]>0)
a[j]=1;
}
memset(b,0,sizeof(b));
}
if(a[0]>0)cout<<"Divisible"<<endl;
else cout<<"Not divisible"<<endl;
}
//system("PAUSE");
return 0;
}