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1 /*
2 题意：问区间内x的出现的次数
3 莫队算法：用一个cnt记录x的次数就可以了
4 还有二分查找的方法
5 */
6 #include <cstdio>
7 #include <algorithm>
8 #include <cmath>
9 #include <cstring>
10 using namespace std;
11
12 const int MAXN = 1e5 + 10;
13 const int INF = 0x3f3f3f3f;
14 struct Data
15 {
16 int b, l, r, x;
17 int id;
18 }data[MAXN];
19 int a[MAXN];
20 int cnt[MAXN];
21 int ans[MAXN];
22 int n, q;
23
24 bool cmp(Data x, Data y)
25 {
26 if (x.b == y.b) return x.r < y.r;
27 return x.b < y.b;
28 }
29
30 void Modui(void)
31 {
32 memset (cnt, 0, sizeof (cnt));
33
34 int l = 1, r = 0;
35 for (int i=1; i<=q; ++i)
36 {
37 while (data[i].l < l) cnt[a[--l]]++;
38 while (data[i].l > l) cnt[a[l]]--, l++;
39 while (data[i].r > r) cnt[a[++r]]++;
40 while (data[i].r < r) cnt[a[r]]--, r--;
41
42 ans[data[i].id] = cnt[data[i].x];
43 }
44
45 for (int i=1; i<=q; ++i)
46 {
47 printf ("%d\n", ans[i]);
48 }
49 }
50
51 int main(void) //FZU 2072 Count
52 {
53 // freopen ("A.in", "r", stdin);
54
55 while (scanf ("%d%d", &n, &q) == 2)
56 {
57 int block = (int) sqrt (n * 1.0);
58 for (int i=1; i<=n; ++i) scanf ("%d", &a[i]);
59 for (int i=1; i<=q; ++i)
60 {
61 scanf ("%d%d%d", &data[i].l, &data[i].r, &data[i].x);
62 data[i].b = data[i].l / block; data[i].id = i;
63 }
64
65 sort (data+1, data+1+q, cmp);
66
67 Modui ();
68 }
69
70 return 0;
71 }
1 #include <cstdio>
2 #include <algorithm>
3 #include <cmath>
4 #include <cstring>
5 #include <vector>
6 using namespace std;
7
8 const int MAXN = 1e5 + 10;
9 const int INF = 0x3f3f3f3f;
10 int a[MAXN];
11 int n, q;
12 vector<int> cnt[MAXN];
13
14 int cal(int x, int r)
15 {
16 if (!cnt[x].size ()) return 0;
17 int pos = upper_bound (cnt[x].begin (), cnt[x].end (), r) - cnt[x].begin () - 1;
18 return pos;
19 }
20
21 int main(void) //FZU 2072 Count
22 {
23 // freopen ("A.in", "r", stdin);
24
25 while (scanf ("%d%d", &n, &q) == 2)
26 {
27 for (int i=1; i<=n; ++i)
28 {
29 scanf ("%d", &a[i]);
30 }
31 for (int i=1; i<=n; ++i) cnt[a[i]].clear ();
32 for (int i=1; i<=n; ++i) cnt[a[i]].push_back (i);
33
34 for (int i=1; i<=q; ++i)
35 {
36 int l, r, x; scanf ("%d%d%d", &l, &r, &x);
37 printf ("%d\n", cal (x, r) - cal (x, l - 1));
38 }
39 }
40
41 return 0;
42 }
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原文地址：http://www.cnblogs.com/Running-Time/p/4650189.html
