Codejam Qualification Round 2019

時間  2019-12-08
標籤 codejam qualification 简体版
原文   原文鏈接

本渣清明節 閒裏偷忙 作了一下codejam
水平不出意外的在投稿以後一落千丈
後兩題的hidden test居然都掛了java

A. Foregone Solution
水題，稍微判斷一下特殊狀況(好比1000, 5000這種)就行了python

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cstring>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <algorithm>
#include <functional>
// #include <assert.h>
#include <iomanip>
using namespace std;
const int N = 7005;
const int M = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
typedef long long ll;


char num[105];
char a[105];
char b[105];
int main() {
    int _;
    scanf("%d", &_);
    for(int cas = 1; cas <= _; ++cas) {
        scanf("%s", num);

        bool flag = true;
        for(int i = 0, len = strlen(num); i < len; ++i) {
            if(num[i] == '4') {
                a[i] = '2';
                b[i] = '2';
                flag = false;
            } else {
                a[i] = num[i];
                b[i] = '0';
            }
        }

        if(flag == true) {
            int pos = -1;
            
            for(int i = strlen(num) - 1; i >= 0; --i) {
                if(num[i] != '0') {
                    pos = i;
                    // if(num[i] == '1') {
                    //     hasOne = true;
                    // }
                    break;
                }
            }
            int hasOne = false; 
            for(int i = 0, len = strlen(num); i < len; ++i) {
                if(i < pos) {
                    a[i] = num[i];
                    b[i] = '0';
                } else if(i == pos) {
                    if(num[i] == '5') {
                        a[i] = '3';
                        b[i] = '2';
                    } else if(num[i] == '1' && i != strlen(num) - 1) {
                        a[i] = '0';
                        b[i] = '0';
                        hasOne = true;
                    } else {
                        a[i] = num[i] - 1;
                        b[i] = '1';
                    }
                } else {
                    if(hasOne) {
                        a[i] = '9';
                        b[i] = (i == strlen(num) - 1) ? '1' :'0';
                    } else {
                        a[i] = num[i];
                        b[i] = '0';
                    }
                }
            }
        }

        printf("Case #%d: ", cas);
        for(int i = 0, len = strlen(num), flag = true; i < len; ++i) {
            if(a[i] == '0' && flag) {
            } else {
                flag = false;
                printf("%c", a[i]);
            }
        }
        printf(" ");
        for(int i = 0, len = strlen(num), flag = true; i < len; ++i) {
            if(b[i] == '0' && flag) {
            } else {
                flag = false;
                printf("%c", b[i]);
            }
        }
        printf("\n");
    
    }
    return 0;
}

B. You Can Go Your Own Way
水題，直接和她反着來就行了ios

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cstring>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <algorithm>
#include <functional>
// #include <assert.h>
#include <iomanip>
using namespace std;
const int N = 7005;
const int M = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
typedef long long ll;

char seq[100005];
char result[100005];
int main() {
    int _;
    scanf("%d", &_);
    for(int cas = 1; cas <= _; ++cas) {
        int n;
        scanf("%d %s", &n, seq);
        for(int i = 0, len = strlen(seq); i < len; ++i) {
            if(seq[i] == 'S')
                result[i] = 'E';
            else 
                result[i] = 'S';
        }
  

        printf("Case #%d: ", cas);
        for(int i = 0, len = strlen(seq); i < len; ++i) {
            printf("%c", result[i]);
        }
        printf("\n");

    
    }
    return 0;
}

C. Cryptopangrams
你們都知道求下相鄰的數的gcd，而後去重，排序
有個小坑就是，序列當中可能相鄰的數是相同的，也就意味着有原始密碼有a,b,a這種。須要找到一個相鄰的數不相同的，而後反推其餘的全部
其次就是大數，c++模版很差找，不如寫java, pythonc++

import java.io.*;
import java.util.*;
import java.math.*;


public class Solution {
    Scanner in = new Scanner(new BufferedInputStream(System.in));
    PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    BigInteger seq[] = new BigInteger[105];
    BigInteger prime[] = new BigInteger[105];
    //    BigInteger num[] = new BigInteger[105];
    int lower_bound(BigInteger[] nums, int begin, int end, BigInteger value) {
        while (begin < end) {
            int mid = begin + (end - begin) / 2;
            if (nums[mid].compareTo(value) == -1) {
                begin = mid + 1;
            } else {
                end = mid;
            }
        }
        return begin;
    }
    void solve() {
        int casNum;
        casNum = in.nextInt();
        for(int cas = 1; cas <= casNum; ++cas) {
            BigInteger n;
            int l;
            n = in.nextBigInteger(); l = in.nextInt();
            for(int i = 0; i < l; ++i) {
                seq[i] = in.nextBigInteger();
            }

            int pos = -1;
            for(int i = 1; i < l; ++i) {
                if(seq[i].equals(seq[i-1]) == false) {
                    prime[i] = seq[i].gcd(seq[i-1]);
                    pos = i;
                    break;
                }
            }

            for(int i = pos - 1; i >= 0; --i) {
                prime[i] = seq[i].divide(prime[i + 1]);
            }
            for(int i = pos + 1; i <= l; ++i) {
                prime[i] = seq[i-1].divide(prime[i-1]);
            }

            List<BigInteger> num=new ArrayList<>();
            for(int i = 0; i <= l; ++i) {
                num.add(prime[i]);
            }
            Set<BigInteger> uniqueGas = new HashSet<BigInteger>(num);
//            out.println(uniqueGas.size());
//            for(BigInteger i : uniqueGas) {
//                out.println(i);
//            }
            BigInteger[] result = uniqueGas.toArray(new BigInteger[0]);
            Arrays.sort(result, 0, uniqueGas.size());
            assert(uniqueGas.size() == 26);
            // printf("%d\n", (int)num.size());
            // for(int i = 0, len = num.size(); i < len; ++i) printf("%lld ", num[i]); printf("\n");
            out.printf("Case #%d: ", cas);

            for(int i = 0; i <= l; ++i) {
                int tt = lower_bound(result,0, uniqueGas.size(), prime[i]);
                out.printf("%c", tt + 'A');
            }
            out.printf("\n");
        }
        out.flush();
    }
    public static void main(String args[]) {
        new Solution ().solve();

    }
}

D. Dat Bae
若是B沒有小於等於15的限制的話，本渣以爲須要10次查詢，這也是本渣的解法
不斷二分01查詢，好比0011, 0101這樣查，就能最後判斷出來每一個位置在不在ide

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cstring>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <algorithm>
#include <functional>
#include <assert.h>
#include <iomanip>
using namespace std;
const int N = 7005;
const int M = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
typedef long long ll;

vector<pair<int, int> > prepare;
vector<pair<int, int> > solve; 
vector<int> preClip;
vector<int> solveClip;
int result[2005];
char answer[2005];
int main() {
    int _;
    scanf("%d", &_);
    for(int cas = 1; cas <= _; ++cas) {
        int n, b, f;
        scanf("%d %d %d", &n, &b, &f);
        prepare.clear();
        preClip.clear();
        prepare.push_back(make_pair(1, n));
        preClip.push_back(n - b);

        while(1) {
            solveClip.clear();
            solve.clear();
            int flag = false;
            int maxx = -1;
            for(int i = 0, len = prepare.size(); i < len; ++i) {
                int start = prepare[i].first; int end = prepare[i].second;
                // if(end == start) {
                //     solve.push_back(make_pair(start, end));
                // } else {
                    int mid = (start + end) / 2;
                    solve.push_back(make_pair(start, mid));
                    solve.push_back(make_pair(mid + 1, end));
                // }
                maxx = max(maxx, end - start);
            }
            if(maxx == 1) flag = true;

            for(int i = 0, len = solve.size(); i < len; ++i) {
                int start = solve[i].first; int end = solve[i].second;
                for(int j = start; j <= end; ++j) {
                    printf("%d", (i % 2 == 0) ? 0 : 1);
                }
            }
            printf("\n");
            fflush(stdout);
            scanf("%s", answer);
            for(int i = 0, len = preClip.size(), pre = 0; i < len; ++i) {
                int pos = pre + preClip[i];
                for(int j = pre, endJ = pre + preClip[i]; j < endJ; ++j) {
                    if(answer[j] == '1') {
                        pos = j;
                        break;
                    }
                }
                solveClip.push_back(pos - pre);
                solveClip.push_back(preClip[i] - pos + pre);
                pre += preClip[i];
            }

            if(flag == true) break;

            // assert(solve.size() == solveClip.size());
            // for(int i = 0, len = solve.size(); i < len; ++i) {
            //     int start = solve[i].first; int end = solve[i].second; int val = solveClip[i];
            //     printf("%d %d %d: ", start, end, val);
            // } printf("\n");

            prepare.swap(solve);
            preClip.swap(solveClip);
        }

        assert(solve.size() == solveClip.size());

        for(int i = 0, len = solve.size(); i < len; ++i) {
            int start = solve[i].first; int end = solve[i].second; int val = solveClip[i];
            if(start == end) {
                if(val == 1) {
                    result[start] = 1;
                } else {
                    result[start] = 0;
                }
            }
        }
        for(int i = 1, fir = true; i <= n; ++i) {
            if(result[i] == 0) {
                if(fir) fir = false;
                else printf(" ");
                printf("%d", i - 1);
            } 
        }
        printf("\n");
        fflush(stdout);

        int basa;
        scanf("%d", &basa);
    }
    return 0;
}

/*
1000
5 2 10
answer(0 3)
2 1 10
answer(0)



*/

可是若是隻是5次查詢，這樣不行。反過來經過1，2，4，8爲界的查詢方式，判斷一個16個數的組的borken狀況。
以下所示
0101010101010101
0011001100110011
0000111100001111
0000000011111111
能夠每16個數一個組，由於咱們知道最多15個broken，每組必會有數保留，那麼咱們就能夠判斷出每一個數是在哪一個組當中的。貼下別人的代碼spa

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cstring>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <algorithm>
#include <functional>
#include <assert.h>
#include <iomanip>
using namespace std;
const int N = 7005;
const int M = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
typedef long long ll;

void solve(){
    int F = 4;
    int n, b, fake_f;
    cin >> n >> b >> fake_f;
    // assert(fake_f >= F);
    for(int f = 0; f < F; f++){
        string g;
        for(int i = 0; i < n; i++){
            g += (char)('0' + ((i >> f) & 1));
        }
        cout << g << '\n' << flush;
    }
    vector<int> answers(n-b);
    for(int f = 0; f < F; f++){
        string res;
        cin >> res;
        for(int i = 0; i < n-b; i++){
            answers[i] ^= (res[i] - '0') << f;
        }
        // for(int i = 0; i < n - b; ++i) {
        //     printf("%d ", answers[i]);
        // } printf("\n");
    }
    vector<int> broken;
    int z = 0;
    for(int i = 0; i < n-b; i++){
        while((z & 15) != answers[i]){
            cout << z << ' ';
            z++;
        }
        z++;
    }
    while(z < n){
        cout << z << ' ';
        z++;
    }
    cout << '\n';
    cout << flush;
    int res;
    cin >> res;
}


int main(){
    int T;
    cin >> T;
    for(int t = 1; t <= T; t++){
        solve();
    }
}

/*
1000
5 2 10

01010
00110
00001
00000
00000
100
010
001
000
000
answer(0 3)
000
010
001
000
000
answer(0 3)

2 1 10
answer(0)



*/
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