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1 /*
2 设一个b[]来保存每一个a[]的质因数的id，从后往前每一次更新质因数的id，
3 若没有，默认加0，nlogn复杂度；
4 我用暴力竟然水过去了:)
5 */
6 #include <cstdio>
7 #include <iostream>
8 #include <cstring>
9 #include <string>
10 #include <algorithm>
11 using namespace std;
12
13 const int MAXN = 1e4 + 10;
14 const int INF = 0x3f3f3f3f;
15 int a[MAXN], b[MAXN];
16
17 int main(void) //BestCoder Round #39 1002 Mutiple
18 {
19 //freopen ("1002.in", "r", stdin);
20
21 int n;
22
23 while (scanf ("%d", &n) == 1)
24 {
25 memset (b, 0, sizeof (b));
26 for (int i=1; i<=n; ++i) scanf ("%d", &a[i]);
27
28 long long ans = 0;
29 for (int i=n; i>=1; --i)
30 {
31 ans += b[a[i]];
32 for (int j=1; j*j<=a[i]; ++j)
33 {
34 if (a[i] % j == 0)
35 {
36 b[j] = i; b[a[i]/j] = i;
37 }
38 }
39 }
40
41 printf ("%lld\n", ans);
42 }
43
44 return 0;
45 }
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <string> 5 #include <algorithm> 6 using namespace std; 7 8 const int MAXN = 1e4 + 10; 9 const int INF = 0x3f3f3f3f; 10 int a[MAXN]; 11 12 int main(void) //BestCoder Round #39 1002 Mutiple 13 { 14 //freopen ("1002.in", "r", stdin); 15 16 int n; 17 long long sum = 0; 18 19 while (scanf ("%d", &n) == 1) 20 { 21 for (int i=1; i<=n; ++i) scanf ("%d", &a[i]); 22 23 sum = 0; int k = 0; 24 for (int i=1; i<=n-1; ++i) 25 { 26 k = 0; 27 for (int j=i+1; j<=n; ++j) 28 { 29 if (a[j] % a[i] == 0) 30 { 31 k = j; break; 32 } 33 } 34 sum += k; 35 } 36 37 printf ("%lld\n", sum); 38 } 39 40 return 0; 41 }
暴力+降复杂度 BestCoder Round #39 1002 Mutiple
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原文地址：http://www.cnblogs.com/Running-Time/p/4460800.html
