线段树分割 简单思维题

https://acm.hdu.edu.cn/showproblem.php?pid=6983

杭电多校的一道题，深刻觉得自己太菜了。
开始打表，然后找结构变化，写了一大篇，乱的不行。
标程是记忆化搜索，醉了，话说也好久没用过了。

乱写ac代码：

#include<bits/stdc++.h>

using namespace std;
#define  int long long
#define rep(i, x, y) for(auto i=(x);i<=(y);++i)
#define dep(i, x, y) for(auto i=(x);i>=(y);--i)
#define gcd(a, b) __gcd(a,b)
const long long mod = 1e9 + 7;
const int maxn = 100000 + 5;
int lowbit(int x) { return x & -x; }

bool ispow(int n) { return (n & (n - 1)) == 0; }//O(1) 判断是否是 2^k（2的k次方）
int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}

int n, k;
int sum=0;

void build(long long l, long long r) {
    sum++;
    if (r - l + 1 <= k) return ;
    long long mid = (l + r) / 2;
     build(l, mid);
     build(mid + 1, r);
    return ;
}
int build1(int l,int r,int z,int cnt) {
    if (r - l + 1 <= k) return cnt;
    long long mid = (l + r) / 2;
    cnt=build1(1,mid,z<<1,cnt+1);
    return cnt;
}

int f(int x){
    int pp= build1(1, x, 1, 0);
    if(ispow(pp)||ispow(pp-1))
        return 2*pp-1;
    return ((long long)1<< build1(1, x, 1, 0))+f(x/2);
}
signed main() {

  //freopen("1.in","r",stdin);
    int t ;

    cin>>t;
    while(t--){
        cin >> n >> k;
        sum = 0;
       // build(1, n);
        //cout << sum << endl;
        int cnt=0;
        int ans=0;
        int ps=0;
        int N=n;
        if (n % k > 0)n = n / k + 1;
        else n = n / k;
       // if(!ispow(n))
        for(int i=1;i<=60;i++){
            if(n>((long long)1<<(i-1))&&n<=((long long)1<<(i)))
            {
                cnt=i-1;
                ps = n - ((long long) 1 << cnt);
                break;
            }
        }
        N-=((int)1<<cnt)*k;
        N*=2;
        if(N>(int)1<<(cnt+1))N=( (int) 1 << (cnt + 1));

        int flag=0;
        ans=N;
        n=((int)1<<cnt);
        ans+=n;
        while(n!=1){
           flag=n%2;
           ans+=n/2;
           n=n/2+flag;
        }
        cout<<ans<<endl;

    }
    return 0;
}

记忆化搜索：

#include<bits/stdc++.h>

using namespace std;
#define  int long long
#define ll long long
#define rep(i, x, y) for(auto i=(x);i<=(y);++i)
#define dep(i, x, y) for(auto i=(x);i>=(y);--i)
#define gcd(a, b) __gcd(a,b)
const long long mod = 1e9 + 7;
const int maxn = 100000 + 5;
int lowbit(int x) { return x & -x; }

bool ispow(int n) { return (n & (n - 1)) == 0; }//O(1) 判断是否是 2^k（2的k次方）
int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}

int Case;
ll n, k;
map<int ,int >mp;

ll dfs(ll n) {
    int res=0;
    if(mp[n])return mp[n];
    if(k>=n)
        return 1;
    mp[n]=dfs(n/2)+dfs((n+1)/2)+1;
    res+=mp[n];
    return res;

}
signed main() {
    scanf("%d", &Case);
    while (Case--) {
        scanf("%lld%lld", &n, &k);
        mp.clear();
        printf("%lld\n", dfs(n));
    }
}