2011-06-01 22:23:39Morris
a064. SPOJ 4580.ABCDEF
http://zerojudge.tw/ShowProblem?problemid=a064
內容 ：
You are given a set S of integers between -30000 and 30000 (inclusive).
Find the total number of sextuples that satisfy:
輸入說明 ：
The first line contains integer N (1 ≤ N ≤ 100), the size of a set S.
Elements of S are given in the next N lines, one integer per line. Given numbers will be distinct.
輸出說明 ：
Output the total number of plausible sextuples.
範例輸入 ：
112232-1135710
範例輸出 ：
142410
提示 ：
| Added by: | Luka Kalinovcic |
| Date: | 2009-07-13 |
| Time limit: | 2s |
| Source limit: | 50000B |
| Languages: | All except: ERL JS PERL 6 |
| Resource: | own problem |
题目来源SPOJ 4580，输入只有一笔，不用EOF判断结束。
出處 ：
/**********************************************************************************/
/* Problem: a064 "SPOJ 4580.ABCDEF" from SPOJ 4580 */
/* Language: C */
/* Result: AC (780ms, 737KB) on ZeroJudge */
/* Author: morris1028 at 2011-05-31 22:36:21 */
/**********************************************************************************/
#include<stdio.h>
#include<stdlib.h>
#define move 60000
main() {
int N, S[100], i, j, k, l;
while(scanf("%d", &N) == 1) {
for(i = 0; i < N; i++)
scanf("%d", &S[i]);
int ef[120001] = {};
long long A = 0;
for(i = 0; i < N; i++)
for(j = 0; j < N; j++)
ef[(S[i]+S[j]+move)]++;
for(i = 0; i < N; i++) {
for(j = i+1; j < N; j++)
for(l = 0; l < N; l++)
for(k = 0; k < N; k++){
if(S[k] != 0 && (S[i]*S[j]+S[l])%S[k] == 0 && abs((S[i]*S[j]+S[l])/S[k]) < 60000 )
A+=2*ef[(S[i]*S[j]+S[l])/S[k]+move];
}
j = i;
for(l = 0; l < N; l++)
for(k = 0; k < N; k++){
if(S[k] != 0 && (S[i]*S[j]+S[l])%S[k] == 0 && abs((S[i]*S[j]+S[l])/S[k]) < 60000 )
A+=ef[(S[i]*S[j]+S[l])/S[k]+move];
}
}
printf("%lld\n", A);
}
return 0;
}
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