HDOJ 5567 sequence1



sequence1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 31    Accepted Submission(s): 29

Problem Description

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.

 

Input

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)

 

Output

For each cases, please output an integer in a line as the answer.

 

Sample Input

3 3 21 2 33 3 11 2 3

 

Sample Output

12

 

题意：给定长度为n的序列的序列a,求有多少对i,j(i<j)，使得|ai-aj| mod b ==c。

水题，今天也就过了这一题，膝盖碎了。QAQ~~~

代码如下：

#include<cstdio>#include<cstring>#include<cstdlib>int a[110];int main(){int n,i,j,b,c,ans;while(scanf("%d%d%d",&n,&b,&c)!=EOF){for(i=0;i<n;++i)scanf("%d",&a[i]);ans=0;for(i=0;i<n;++i){for(j=i+1;j<n;++j){if(abs(a[i]-a[j])%b==c)ans++;}}printf("%d\n",ans);}return 0;}