sequence1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 42 Accepted Submission(s): 38
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
Source
题解：水，刚开始还以为要判断数据重复的情况呐，最后发现根本不用，是 how many pairs (i,j) ，指的是i,j的对数。。。
题意就是找 how many pairs (i,j) ( i < j ) for abs(ai−aj) mod b=c.
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define mem(x,y) memset(x,y,sizeof(x))
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
typedef long long LL;
LL m[110];
bool judge(int a,int c,int b){
if((a-c)%b==0)return true;
else return false;
}
int main(){
int n,b,c;
while(~scanf("%d%d%d",&n,&b,&c)){
for(int i=0;i<n;i++)scanf("%I64d",m+i);
LL ans=0;
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(judge(abs(m[i]-m[j]),c,b))ans++;
}
}
printf("%I64d
",ans);
}
return 0;
}