hdoj--5567--sequence1(水题)
2015-11-29 21:32
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sequence1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 345 Accepted Submission(s): 254
[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)
[align=left]Recommend[/align]
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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; }
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