HDOJ 5567 sequence1 (暴力)

最新推荐文章于 2017-12-10 10:44:34 发布

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本文介绍了一个关于寻找数组中满足特定条件的数对数量的问题，通过给出的示例和代码，展示了如何利用双重循环来遍历并判断每一对元素是否符合要求，最终输出符合条件的数对总数。

sequence1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 219 Accepted Submission(s): 168

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

Sample Output

1
2

题意：查看|(a[i]-a[j])|%b==c的组合有多少，i<j

思路：数据只有1k，直接暴力

ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 10000001
#define fab(a) (a)>0?(a):(-a)
#define INF 0xfffffff
#define LL long long
using namespace std;
LL a[MAXN];
int main()
{
    LL n,b,c;
    int i,j;
    while(~scanf("%lld%lld%lld",&n,&b,&c))
    {
        for(i=0;i<n;i++)
        scanf("%lld",&a[i]);
        LL ans=0;
        for(i=0;i<n;i++)
        {
            for(j=i+1;j<n;j++)
            {
                LL k=a[i]-a[j];
                if(k<0) k=-k;
                if(k%b==c)
                ans++;
            }
        }
        printf("%lld\n",ans);
    } 
    return 0;
}