2023.3.10

这几天讲的倍增捏（一开始以优化 dp 为主，现在几乎纯倍增，6）。然后发现我不会快速幂。

贴一个板子。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll ans = 1;
ll a, b, p;
ll aa, bb, mod;
int main()
{
    scanf("%lld%lld%lld", &a, &b, &p);
    aa = a;
    bb = b;
    mod = p;
    while(b)
    {
        if(b % 2) ans = ans * a % p;
        a = a * a % p;
        b /= 2;
    }
    printf("%lld", ans % p);
    return 0;
}

---

3.09 模拟赛 T2，洛谷：P3509 ZAB-Frog

正解是倍增 + 单调队列。但是对于倍增最后空间会卡，所以要开动态数组，这里没有太理解，想明白了再补过来思路。先存一下码。

#include<bits/stdc++.h>
#define ll long long 
using namespace std;
const int N = 1000010;
ll m;
int n, k;
ll a[N], nex[N], f[N], ans[N];
int main()
{
    scanf("%d%d%lld", &n, &k, &m);
    for(int i = 1; i <= n; i ++ ) scanf("%lld", &a[i]);
    int head = 1, tail = k + 1;
    nex[1] = k + 1;
    for(int i = 2; i <= n; i ++ )
    {
        while(tail + 1 <= n && a[i] - a[head] > a[tail + 1] - a[i]) head ++, tail ++;
        if(a[i] - a[head] >= a[tail] - a[i]) nex[i] = head;
        else nex[i] = tail;
    }
    for(int i = 1; i <= n; i ++ ) ans[i] = i;
    
    while(m > 0)
    {
        if(m % 2 == 1) for(int i = 1; i <= n; i ++ )ans[i] = nex[ans[i]];
        m /= 2;
        for(int i = 1; i <= n; i ++ ) f[i] = nex[i];
        for(int i = 1; i <= n; i ++ ) nex[i] = f[f[i]];
    }
    for(int i = 1; i <= n; i ++ ) printf("%lld ", ans[i]);
    
    return 0;
}