本渣清明节 闲里偷忙 做了一下codejam
水平不出意外的在投稿之后一落千丈
后两题的hidden test竟然都挂了

A. Foregone Solution
水题，稍微判断一下特殊情况(比如1000, 5000这种)就好了

#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
水题，直接和她反着来就好了

#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
大家都知道求下相邻的数的gcd，然后去重，排序
有个小坑就是，序列当中可能相邻的数是相同的，也就意味着有原始密码有a,b,a这种。需要找到一个相邻的数不相同的，然后反推其他的所有
其次就是大数，c++模版不好找，不如写java, python

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
如果B没有小于等于15的限制的话，本渣觉得需要10次查询，这也是本渣的解法
不断二分01查询，比如0011, 0101这样查，就能最后判断出来每个位置在不在

#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)*/

但是如果只是5次查询，这样不行。反过来通过1，2，4，8为界的查询方式，判断一个16个数的组的borken情况。
如下所示
0101010101010101
0011001100110011
0000111100001111
0000000011111111
可以每16个数一个组，因为我们知道最多15个broken，每组必会有数保留，那么我们就可以判断出每个数是在哪个组当中的。贴下别人的代码

#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 1001010
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)*/