sequence1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 18 Accepted Submission(s): 17
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)
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 2 1 2 3 3 3 1 1 2 3
Sample Output
1 2
很水的题目,签到吧。
AC代码:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define MAXN 100
#define MAXM 1010
#define eps 1e-8
#define LL long long
using namespace std;
LL b, c;
LL a[110];
bool judge(LL x, LL y){
return abs(x - y) % b == c;
}
int main()
{
int n;
while(scanf("%d%lld%lld", &n, &b, &c) != EOF)
{
int ans = 0;
for(int i = 0; i < n; i++)
scanf("%lld", &a[i]);
for(int i = 0; i < n; i++)
for(int j = i+1; j < n; j++)
if(judge(a[i], a[j]))
ans++;
printf("%d\n", ans);
}
return 0;
}