hdoj--5567--sequence1(水题)

sequence1

[align=center]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 345    Accepted Submission(s): 254
[/align]

[align=left]Problem Description[/align]
Given an array a
with length n,
could you tell me how many pairs (i,j)
( i < j ) for abs(ai−aj) mod b=c.
 

[align=left]Input[/align]
Several test cases(about
5)

For each cases, first come 3 integers, n,b,c(1≤n≤100,0≤c<b≤109)

Then follows n
integers ai(0≤ai≤109)

 

[align=left]Output[/align]
For each cases, please output an integer in a line as the answer.
 

[align=left]Sample Input[/align]

3 3 2
1 2 3
3 3 1
1 2 3

 

[align=left]Sample Output[/align]

1
2

 

[align=left]Source[/align]
BestCoder Round #63 (div.2)

 

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#include<stdio.h>
#include<string.h>
#include<math.h>
int n,b,c;
long long a[1010];
bool judge(long long x,long long y)
{
if(abs(x-y)%b==c)
return true;
return false;
}
int main()
{
while(scanf("%d%d%d",&n,&b,&c)!=EOF)
{
memset(a,0,sizeof(a));
for(int i=0;i<n;i++)
scanf("%lld",&a[i]);
int sum=0;
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
if(judge(a[i],a[j]))
sum++;
}
}
printf("%d\n",sum);
}
return 0;
}