模板 ST表

Posted 2021-01-07 luv-letters

 

某dalao的代码

 

 1 void ST(int n) {
 2     for (int i = 1; i <= n; i++)
 3         dp[i][0] = A[i];
 4     for (int j = 1; (1 << j) <= n; j++) {
 5         for (int i = 1; i + (1 << j) - 1 <= n; i++) {
 6             dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
 7         }
 8     }
 9 }
10 int RMQ(int l, int r) {
11     int k = 0;
12     while ((1 << (k + 1)) <= r - l + 1) k++;
13     return max(dp[l][k], dp[r - (1 << k) + 1][k]);
14 }

 

我莫名WA的题

 

 1 #include<bits/stdc++.h>
 2 #define maxn 100010 
 3 using namespace std;
 4 int stmin[maxn][20];
 5 int stmax[maxn][20];
 6 int n,m;
 7 void st(){
 8     for(int j=1;(1<<j)<=n;j++)
 9         for(int i=1;i+(1<<i)-1 <= n ;i++){
10             stmin[i][j]    =min(stmin[i][j-1],stmin[i+(1<<(j-1))][j-1]);
11             stmax[i][j]    =min(stmax[i][j-1],stmax[i+(1<<(j-1))][j-1]);    
12         }
13 }
14 int rmq_max (int l,int r){
15     int k=0;
16     while(( 1<<(k+1) ) <=(r-l+1) ) k++;
17     return max(stmax[l][k],stmax[r-(1<<k)+1][k]);
18 }
19 int rmq_min (int l,int r){
20     int k=0;
21     while(( 1<<(k+1) ) <=(r-l+1) ) k++;
22     return min(stmin[l][k],stmin[r-(1<<k)+1][k]);
23 }
24 int main(void){
25     cin>>n>>m;
26     for(int i=1;i<=n;i++)  cin>>stmin[i][0];
27     st();
28     for(int i=1;i<m-1;i++)   cout<<0<<endl; 
29     for(int i=m;i<=n+1;i++){
30         cout<<rmq_min(i-m,i)<<endl;
31     }
32 } 

 

困得不行，可能是个傻逼错误qwq

 

改天再改