6983 杭电多校(2021“MINIEYE杯”中国大学生算法设计超级联赛3) [记忆化搜索]

Posted 2021-08-31 你可以当我没名字

Segment Tree with Pruning

*Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 544 Accepted Submission(s): 404
*

Problem Description

Chenjb is struggling with data stucture now. He is trying to solve a problem using segment tree. Chenjb is a freshman in programming contest, and he wrote down the following C/C++ code and ran ‘‘Node* root = build(1, n)’’ to build a standard segment tree on range [1,n]:

Node* build(long long l, long long r) {
    Node* x = new(Node);
    if (l == r) return x;
    long long mid = (l + r) / 2;
    x -> lchild = build(l, mid);
    x -> rchild = build(mid + 1, r);
    return x;
}

Chenjb submitted his code, but unfortunately, got MLE (Memory Limit Exceeded). Soon Chenjb realized that his program will new a large quantity of nodes, and he decided to reduce the number of nodes by pruning:

Node* build(long long l, long long r) {
    Node* x = new(Node);
    if (r - l + 1 <= k) return x;
    long long mid = (l + r) / 2;
    x -> lchild = build(l, mid);
    x -> rchild = build(mid + 1, r);
    return x;
}

You know, Chenjb is a freshman, so he will try different values of k to find the optimal one. You will be given the values of n and k, please tell him the number of nodes that will be generated by his new program.

Input

The first line contains a single integer T (1≤T≤10000), the number of test cases. For each test case:

The only line contains two integers n and k (1≤k≤n≤1018), denoting a query.

Output

For each query, print a single line containing an integer, denoting the number of segment tree nodes.

Sample Input

3
100000 1
100000 50
1000000000000000000 1

Sample Output

199999
4095
1999999999999999999

Source

2021“MINIEYE杯”中国大学生算法设计超级联赛（3）

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代码与解释

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

看懂题目,会记忆化搜索就能过了

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <iostream>
#include <sstream>
#include <set>
#include <map>
#include <queue>
#include <bitset>
#include <vector>
#include <limits.h>
#include <assert.h>
#include <functional>
#include <numeric>
#include <ctime>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define pb          push_back
#define ppb         pop_back
#define lbnd        lower_bound
#define ubnd        upper_bound
#define endl        '\\n'
#define mll         map<ll,ll>
#define msl         map<string,ll>
#define mls         map<ll, string>
#define rep(i,a,b)  for(ll i=a;i<b;i++)
#define repr(i,a,b) for(ll i=b-1;i>=a;i--)
#define trav(a, x)  for(auto& a : x)
#define pll         pair<ll,ll>
#define vl          vector<ll>
#define vll         vector<pair<ll, ll>>
#define vs          vector<string>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define DEBUG       cerr<<"/n>>>I'm Here<<</n"<<endl;
#define display(x)  trav(a,x) cout<<a<<" ";cout<<endl;
#define what_is(x)  cerr << #x << " is " << x << endl;
#define ini(a)      memset(a,0,sizeof(a))
#define ini2(a,b)   memset(a,b,sizeof(a))
#define rep(i,a,b)  for(int i=a;i<=b;i++)
#define sc          ll T;cin>>T;for(ll Q=1;Q<=T;Q++)
#define lowbit(x)   x&(-x)
#define pr          printf
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define DBG(x) \\
    (void)(cout << "L" << __LINE__ \\
    << ": " << #x << " = " << (x) << '\\n')
#define TIE \\
    cin.tie(0);cout.tie(0);\\
    ios::sync_with_stdio(false);
//#define long long int

//using namespace __gnu_pbds;

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int maxn = 100009;
const ll N = 1000000007;

mll ma;
ll n, k, cnt=0;

ll build(ll l, ll r){
    if(ma.count(r-l))return ma[r-l];//存在 
    if(r-l+1<=k){
        return 1;
    }
    ll mid = (l+r)/2, tmp = 1;
    tmp += build(l,mid);
    tmp += build(mid+1,r);
    return ma[r-l] = tmp;
}

void solve(){
    ma.clear();
    cin>>n>>k;
    cout<<build(1,n)<<"\\n";
}


int main()
{
//	TIE;
//    #ifndef ONLINE_JUDGE
//    freopen ("input.txt","r",stdin);
//    #else
//    #endif
//	solve();
    sc{solve();}
//    sc{cout<<"Case "<<Q<<":"<<endl;solve();}
}