算法课程笔记——如何倍增

快速幂

读入量大于1e5不要用cin读入，要用也要关闭同步流

第i个o次方的父亲

#include<bits/stdc++.h>usingnamespacestd;
#definemaxn 110000#definell long longintn, a[maxn], f[maxn][40];
intquery(intl, intr){
    intk = (int)(log((r - l + 1) * 1.0) / log(2.0));
    returnmin(f[l][k], f[r - (1<< k) + 1][k]);
}

intmain(){
    scanf("%d", &n);
    for(inti = 1; i <= n; ++ i )
        scanf("%d", &a[i]), f[i][0] = a[i];
    for(intj = 1; j <= (int)(log(n * 1.0) / log(2.0)); ++ j )
        for(inti = 1; i + (1<< j) - 1<= n; ++ i )
            f[i][j] = min(f[i][j - 1], f[i + (1<< (j - 1))][j - 1]);
    intq;
    scanf("%d", q);
    while(q -- )
    {
        intl, r;
        scanf("%d%d", &l, &r);
        printf("%d\n", query(l, r));
    }
   
    return0;
}

#include<bits/stdc++.h>usingnamespacestd;
#definemaxn 110000#definell long longintn, a[maxn], f[maxn][40];
intquery(intl, intr){
    intk = (int)(log((r - l + 1) * 1.0) / log(2.0));
    returnmin(f[l][k], f[r - (1<< k) + 1][k]);
}

intmain(){
    scanf("%d", &n);
    for(inti = 1; i <= n; ++ i )
        scanf("%d", &a[i]), f[i][0] = a[i];
    for(intj = 1; j <= (int)(log(n * 1.0) / log(2.0)); ++ j )
        for(inti = 1; i + (1<< j) - 1<= n; ++ i )
            f[i][j] = min(f[i][j - 1], f[i + (1<< (j - 1))][j - 1]);
    intq;
    scanf("%d", q);
    while(q -- )
    {
        intl, r;
        scanf("%d%d", &l, &r);
        printf("%d\n", query(l, r));
    }
   
    return0;
}