蓝桥杯 - 连号区间数

 1 // 两遍RMQ写法
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 const int N = 50000 + 10;
 5 int n, a[N], ma[N], mi[N], d1[N][16], d2[N][16];
 6 
 7 void RMQ_INIT1(){
 8     for(int i = 0; i < n; i++) d1[i][0] = a[i];
 9     for(int j = 1; (1<<j) <= n; j++)
10         for(int i = 0; i + (1<<j) -1 < n; i++)
11             d1[i][j] = min(d1[i][j-1], d1[i+(1<<(j-1))][j-1]);
12 }
13 
14 void RMQ_INIT2(){
15     for(int i = 0; i < n; i++) d2[i][0] = a[i];
16     for(int j = 1; (1<<j) <= n; j++)
17         for(int i = 0; i + (1<<j) -1 < n; i++)
18             d2[i][j] = max(d2[i][j-1], d2[i+(1<<(j-1))][j-1]);
19 }
20 
21 int query1(int L, int R){
22     int k = 0;
23     while((1<<(k+1)) <= R-L+1) k++;
24     return min(d1[L][k], d1[R-(1<<k)+1][k]);
25 }
26 
27 int query2(int L, int R){
28     int k = 0;
29     while((1<<(k+1)) <= R-L+1) k++;
30     return max(d2[L][k], d2[R-(1<<k)+1][k]);
31 }
32 
33 int main(){
34     scanf("%d", &n);
35     for(int i = 0; i < n; i++) scanf("%d", a+i);
36 
37     RMQ_INIT1();
38     RMQ_INIT2();
39     int ans = 0;
40     for(int i = 0; i < n-1; i++){
41         for(int j = i+1; j < n; j++){
42             int mini = query1(i, j), maxi = query2(i, j);
43             if(j - i == maxi - mini) ans++;
44             // printf("%d %d , %d %d\n", i, j, mini, maxi);
45         }
46     }
47     printf("%d", ans+n);
48     return 0;
49 }

 1 // RMQ的非模版结构体写法，传入STL的max和min法
 2 // 注意用到了typedef简化了max和min的定义
 3 // template< class T > 
 4 // const T& max( const T& a, const T& b );
 5 #include<bits/stdc++.h>
 6 using namespace std;
 7 const int N = 50000 + 10;
 8 int n, a[N];
 9 
10 struct RMQ{
11     typedef const int& (&type)(const int&, const int&);
12     type func;
13     int n, **d;
14 
15     RMQ(type f, int size): func(f), n(size){
16         d = new int*[n];
17         for(int i = 0; i < n; i++) d[i] = new int[(int)(log(n)/log(2))+1];
18         RMQ_INIT();
19     }
20 
21     ~RMQ(){
22         delete[] d;
23     }
24 
25     void RMQ_INIT(){
26         for(int i = 0; i < n; i++) d[i][0] = a[i];
27         for(int j = 1; (1<<j) <= n; j++)
28             for(int i = 0; i + (1<<j) - 1 < n; i++)
29                 d[i][j] = func(d[i][j-1], d[i+(1<<(j-1))][j-1]);
30     }
31 
32     int query(int L, int R){
33         int k = 0;
34         while((1<<(k+1)) <= R-L+1) k++;
35         return func(d[L][k], d[R-(1<<k)+1][k]);
36     }
37 };
38 
39 int main(){
40     scanf("%d", &n);
41     for(int i = 0; i < n; i++) scanf("%d", a+i);
42     RMQ MAX(max,n), MIN(min,n);
43     int ans = 0;
44     for(int i = 0; i < n-1; i++){
45         for(int j = i+1; j < n; j++){
46             int mini = MIN.query(i, j), maxi = MAX.query(i, j);
47             if(j - i == maxi - mini) ans++;
48             // printf("%d %d , %d %d\n", i, j, mini, maxi);
49         }
50     }
51     printf("%d", ans+n);
52     return 0;
53 }

 1 // RMQ的模版结构体写法，传入STL的max和min法
 2 // 数组a也应该单独用模版写，不过不能放在全局部分
 3 #include<bits/stdc++.h>
 4 using namespace std;
 5 const int N = 50000 + 10;
 6 int n, a[N];
 7 template<class T>
 8 struct RMQ{
 9     typedef const T& (&type)(const T&, const T&);
10     type func;
11     int n; T **d;
12 
13     RMQ(type f, int size): func(f), n(size){
14         d = new T*[n];
15         for(int i = 0; i < n; i++) d[i] = new T[(int)(log(n)/log(2))+1];
16         RMQ_INIT();
17     }
18 
19     ~RMQ(){
20         delete[] d;
21     }
22 
23     void RMQ_INIT(){
24         for(int i = 0; i < n; i++) d[i][0] = a[i];
25         for(int j = 1; (1<<j) <= n; j++)
26             for(int i = 0; i + (1<<j) - 1 < n; i++)
27                 d[i][j] = func(d[i][j-1], d[i+(1<<(j-1))][j-1]);
28     }
29 
30     int query(int L, int R){
31         int k = 0;
32         while((1<<(k+1)) <= R-L+1) k++;
33         return func(d[L][k], d[R-(1<<k)+1][k]);
34     }
35 };
36 
37 int main(){
38     scanf("%d", &n);
39     for(int i = 0; i < n; i++) scanf("%d", a+i);
40     RMQ<int> MAX(max,n), MIN(min,n);
41     int ans = 0;
42     for(int i = 0; i < n-1; i++){
43         for(int j = i+1; j < n; j++){
44             int mini = MIN.query(i, j), maxi = MAX.query(i, j);
45             if(j - i == maxi - mini) ans++;
46             // printf("%d %d , %d %d\n", i, j, mini, maxi);
47         }
48     }
49     printf("%d", ans+n);
50     return 0;
51 }