RMQ

#define LG (17)
#define MX (1<<LG)
#define P2(v) (!(v&(v-1)))
int dp[MX][LG+2],G[MX],XX,O=-1;
void ini(int *A,int n){
    if(!XX++)FT(1,MX)G[k]=O+=P2(k);
    F(n)dp[i][0]=i;
    FT(1,k-(1<<k)+n+1)F(n+1-(1<<k))
        if(A[dp[i][k-1]]<A[dp[i+(1<<(k-1))][k-1]])
            dp[i][k]=dp[i][k-1];
        else dp[i][k]=dp[i+(1<<(k-1))][k-1];
}
int qy(int *A,int L,int R){
    int j(G[R-L+1]);
    if(A[dp[L][j]]<=A[dp[R-(1<<j)+1][j]])
        return A[dp[L][j]];
    return A[dp[R-(1<<j)+1][j]];
}
\end{lstlisting}
Obdobně maxium
\begin{lstlisting}[language=C++]
int dp[MX][LG+2],G[MX],XX,O=-1;
void ini(int *A,int n){
    if(!XX++)FT(1,MX)G[k]=O+=P2(k);
    F(n)dp[i][0]=i;
    FT(1,k-(1<<k)+n+1)F(n+1-(1<<k))
        if(A[dp[i][k-1]]>A[dp[i+(1<<(k-1))][k-1]])
            dp[i][k]=dp[i][k-1];
        else dp[i][k]=dp[i+(1<<(k-1))][k-1];
}
int qy(int *A,int L,int R){
    int j(G[R-L+1]);
    if(A[dp[L][j]]>=A[dp[R-(1<<j)+1][j]])
        return A[dp[L][j]];
    return A[dp[R-(1<<j)+1][j]];
}