1019. General Palindromic Number
题意：求数N在b进制下其序列是否为回文串，并输出其在b进制下的表示。
思路：模拟N在2进制下的表示求法，“除b倒取余”，之后判断是否回文。
1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 int Base[30]; 5 int main() 6 { 7 int n, b; 8 scanf("%d%d", &n, &b); 9 int len = 0, tmp = n,yu=tmp%b; 10 while (tmp / b) 11 { 12 Base[len++] = yu; 13 tmp = tmp / b; 14 yu = tmp % b; 15 } 16 Base[len++] = yu; 17 bool flag = true; 18 for (int i = 0, j = len - 1; i <= j&&flag; i++, j--) 19 { 20 if (Base[i] != Base[j]) flag = false; 21 } 22 if (flag) printf("Yes "); 23 else printf("No "); 24 for (int i = len - 1; i >= 0; i--) 25 { 26 printf("%d", Base[i]); 27 if (i) printf(" "); 28 } 29 printf(" "); 30 return 0; 31 }
1020. Tree Traversals
题意：给出二叉树的后序遍历和中序遍历，求层次遍历。
思路：通过后序遍历和中序遍历建立二叉树，然后对二叉树进行BFS。
1 #include<iostream> 2 #include<cstdio> 3 #include<queue> 4 using namespace std; 5 const int maxn = 40; 6 int postOrder[maxn]; 7 int inOrder[maxn]; 8 int L[maxn],R[maxn]; 9 int treeRoot; 10 queue<int>q; 11 int createBinTree(int pt_pl,int pt_pr,int pt_i,int n) 12 { 13 if (n == 0) return -1; 14 int k = pt_i,curRoot= postOrder[pt_pr]; 15 while (curRoot!= inOrder[k])k++;//在中序中找到当前子树的根 16 L[curRoot] = createBinTree(pt_pl,pt_pl+(k- pt_i)-1,pt_i, k-pt_i); 17 R[curRoot] = createBinTree(pt_pl + (k - pt_i),pt_pr-1,k+1,n - (k-pt_i) - 1); 18 return curRoot; 19 } 20 void printLevelOrder(int root) 21 { 22 q.push(root); 23 while (!q.empty()) 24 { 25 int u = q.front(); 26 q.pop(); 27 if (u == treeRoot) printf("%d", u); 28 else printf(" %d", u); 29 if (L[u] != -1) q.push(L[u]); 30 if (R[u] != -1) q.push(R[u]); 31 } 32 printf(" "); 33 } 34 35 int main() 36 { 37 int n; 38 scanf("%d", &n); 39 for (int i = 0; i < n; i++) scanf("%d", postOrder + i); 40 for (int i = 0; i < n; i++) scanf("%d", inOrder + i); 41 treeRoot = postOrder[n-1]; 42 createBinTree(0,n - 1, 0, n); 43 printLevelOrder(treeRoot); 44 return 0; 45 }
1021. Deepest Root
题意：给出n个点和n-1条边。若是只有一个连通分量，且为无环图，则可将其看成一棵树，求拥有最大深度的树的所有根，并按字典序升序输出。否则输出连通块的数目。
思路：并查集求连通块；DFS判断是否有环(傻了，如果只有1个连通块，则因为边数只有n-1条，只能是无环图)；BFS求树的深度。可通过连续两次BFS求出树的最大直径。
1 #include<iostream> 2 #include<set> 3 #include<queue> 4 #include<cstring> 5 #include<cstdio> 6 using namespace std; 7 const int maxn = 10010; 8 int pre[maxn],cnt_cpt; 9 set<int>ans; 10 int lvl[maxn]; 11 bool vis[maxn]; 12 struct edge 13 { 14 int to,next; 15 edge(int tt=0,int nn=0):to(tt),next(nn){} 16 }; 17 edge Edge[maxn << 1]; 18 int Head[maxn], totedge; 19 void addedge(int from, int to) 20 { 21 Edge[totedge] = edge(to, Head[from]); 22 Head[from] = totedge++; 23 Edge[totedge] = edge(from, Head[to]); 24 Head[to] = totedge++; 25 26 } 27 int max_deep; 28 int Find(int x) 29 { 30 if (pre[x] == x) return x; 31 else 32 { 33 int fa = pre[x]; 34 pre[x] = Find(fa); 35 return pre[x]; 36 } 37 } 38 void Join(int u, int v) 39 { 40 int fu = Find(u), fv = Find(v); 41 if (fu != fv) 42 { 43 if (fv > fu) fv ^= fu, fu ^= fv, fv ^= fu; 44 pre[fu] = fv; 45 cnt_cpt--; 46 } 47 } 48 void BFS(int u) 49 { 50 vis[u] = true; 51 queue<int>q; 52 q.push(u); 53 while (!q.empty()) 54 { 55 int v = q.front(); 56 q.pop(); 57 for (int i = Head[v]; i != -1; i = Edge[i].next) 58 { 59 int t = Edge[i].to; 60 if (!vis[t]) 61 { 62 vis[t] = true; 63 lvl[t] = lvl[v] + 1; 64 if (lvl[t] > max_deep) max_deep = lvl[t]; 65 q.push(t); 66 } 67 } 68 } 69 } 70 //bool DFS(int u,int pre) 71 //{ 72 // vis[u] = true; 73 // for (int i = Head[u]; i != -1; i = Edge[i].next) 74 // { 75 // int v = Edge[i].to; 76 // if (v == pre) continue; 77 // if (vis[v]) return true; 78 // else 79 // { 80 // bool flag = DFS(v, u); 81 // if (flag) return true; 82 // } 83 // } 84 // vis[u] = false; 85 // return false; 86 //} 87 int main() 88 { 89 int n; 90 scanf("%d", &n); 91 92 for (int i = 1; i <= n; i++) pre[i] = i; 93 cnt_cpt = n; 94 totedge = 0; 95 memset(Head, -1, sizeof(Head)); 96 97 for (int i = 1; i < n; i++) 98 { 99 int u, v; 100 scanf("%d%d", &u, &v); 101 Join(u, v); 102 addedge(u, v); 103 } 104 if (cnt_cpt > 1) 105 { 106 printf("Error: %d components ", cnt_cpt); 107 } 108 else 109 { 110 max_deep = 0; 111 BFS(1); 112 int next_root; 113 for (int i = 1; i <= n; i++)if (lvl[i] == max_deep) ans.insert(i), next_root=i; 114 max_deep = 0; 115 memset(lvl, 0, sizeof(lvl)); 116 memset(vis, 0, sizeof(vis)); 117 BFS(next_root); 118 for(int i=1;i<=n;i++)if (lvl[i] == max_deep) ans.insert(i); 119 int sz = ans.size(); 120 set<int>::iterator it = ans.begin(); 121 for (; it != ans.end(); it++) 122 { 123 printf("%d ", *it); 124 } 125 } 126 return 0; 127 }
1022. Digital Library
题意：给出n个图书信息：ID、title、author、key words、publisher、publish year.之后有m个询问，求指定title/author/key word/publisher/publish year的图书的ID。
思路：用map建立title、author、key word、publisher、publish year到ID的映射，用set存放被映射的ID。
一些技巧：sstream的使用、用fgets(str,size,stdin)或scanf("%[^ ]")读入一行带有空格的字符串(注意用getchar()吸收‘ ’).
1 #include<iostream> 2 #include<map> 3 #include<set> 4 #include<string> 5 #include<sstream> 6 #include<cstdio> 7 using namespace std; 8 9 map<string, set<string> >mp[5]; 10 char tmp[90]; 11 string id,t1,t2; 12 int main() 13 { 14 int n; 15 scanf("%d", &n); 16 for (int i = 0; i < n; i++) 17 { 18 scanf("%s", tmp);//ID,fgets(tmp,90,stdin); 19 id = tmp; 20 getchar(); 21 scanf("%[^ ]", tmp);//title 22 mp[0][tmp].insert(id); 23 getchar(); 24 scanf("%[^ ]", tmp);//author 25 mp[1][tmp].insert(id); 26 getchar(); 27 scanf("%[^ ]", tmp);//key words 28 stringstream sss((t1=tmp)); 29 while (sss >> t2) mp[2][t2].insert(id); 30 getchar(); 31 scanf("%[^ ]", tmp);//publisher 32 mp[3][tmp].insert(id); 33 getchar(); 34 scanf("%[^ ]", tmp);//year 35 mp[4][tmp].insert(id); 36 } 37 int m; 38 scanf("%d", &m); 39 for (int i = 1; i <= m; i++) 40 { 41 getchar(); 42 scanf("%[^ ]", tmp); 43 printf("%s ", tmp); 44 set<string>&st = mp[tmp[0] - '1'][tmp + 3]; 45 if (st.size() == 0) printf("Not Found "); 46 else 47 { 48 set<string>::iterator it = st.begin(); 49 for (; it != st.end(); it++) 50 { 51 printf("%s ", (*it).c_str()); 52 } 53 } 54 } 55 return 0; 56 }
1023. Have Fun with Numbers
题意：将给出的一个大数*2，判断其是否可由原来的数通过排列得到。
思路：数组模拟大数*2.
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 char str[30],dst[30]; 6 int main() 7 { 8 scanf("%s", str + 1); 9 int len = strlen(str + 1); 10 int tadd = 0; 11 for (int i = len; i >= 1; i--) 12 { 13 tadd += 2*(str[i] - '0'); 14 dst[i] = '0'+tadd % 10; 15 tadd /= 10; 16 } 17 if (tadd) dst[0] = '0'+tadd; 18 dst[len + 1] = ' '; 19 if (tadd) printf("No %s ",dst); 20 else 21 { 22 bool flag = true; 23 for (int i = 0; i <= 9; i++) 24 { 25 int c1 = 0,c2 = 0; 26 for (int j = 1; j <= len; j++) if (str[j] == '0' + i)c1++; 27 for (int j = 1; j <= len; j++) if (dst[j] == '0' + i) c2++; 28 if (c1 != c2) 29 { 30 printf("No "); 31 flag = false; 32 break; 33 } 34 } 35 if (flag) printf("Yes "); 36 printf("%s ", dst + 1); 37 } 38 return 0; 39 }
1024. Palindromic Number
题意：给出一个<=10^10的数，每次将其颠倒后与原数相加，若结果为回文数，则输出对应的数和颠倒的次数。
思路：用字符串数组模拟大数加法。
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 using namespace std; 7 char init[200], dst[200],ans[200]; 8 void Reverse() 9 { 10 reverse_copy(init, init + strlen(init), dst); 11 return; 12 } 13 void Add() 14 { 15 int len = strlen(init); 16 int pos = len, tmp = 0; 17 for (int i = len - 1; i >= 0; --i) 18 { 19 tmp += init[i] - '0' + dst[i] - '0'; 20 ans[pos--] = '0' + tmp % 10; 21 tmp /= 10; 22 } 23 int st, ed = len; 24 if (tmp) ans[pos] = '0' + tmp, st = 0; 25 else ans[pos] = ' ', st = 1; 26 ans[ed + 1] = ' '; 27 memcpy(init, ans + st, ed - st + 2); 28 } 29 bool isPar() 30 { 31 int len = strlen(init); 32 for (int i = len-1; i >=len/2; i--) 33 { 34 if (init[i] != init[len - i-1]) 35 { 36 return false; 37 } 38 } 39 40 return true; 41 } 42 int main() 43 { 44 long long n, k; 45 scanf("%s%lld", init, &k); 46 long long cur = 0; 47 while (!isPar()) 48 { 49 if (cur == k) break; 50 cur++; 51 Reverse(); 52 Add(); 53 } 54 printf("%s %lld ", init, cur); 55 return 0; 56 }
1025. PAT Ranking
题意：有若干子测试，每个子测试中有ki个人，给出其编号和成绩。现在需要你给出最终的成绩排名，给出每个人的编号、最终排名、子测试序号、子测试排名。
思路：vector应用。用merge对多个vector合并；用assign来对vector赋值。
1 #include<iostream> 2 #include<vector> 3 #include<cstdio> 4 #include<algorithm> 5 using namespace std; 6 struct node 7 { 8 string id; 9 int score; 10 int loc_num; 11 int loc_rank; 12 node(string ii="",int ss=0,int ln=0,int lr=0):id(ii),score(ss),loc_num(ln),loc_rank(lr){} 13 }; 14 vector<node>ans; 15 bool Cmp(node&a, node&b) 16 { 17 if (a.score == b.score) return a.id < b.id; 18 else return a.score > b.score; 19 } 20 char ID[20]; 21 int main() 22 { 23 int n; 24 scanf("%d", &n); 25 int loc_num = 1; 26 while (n--) 27 { 28 int k; 29 scanf("%d", &k); 30 vector<node>tmp; 31 for (int i = 0; i < k; i++) 32 { 33 int score; 34 scanf("%s%d", ID, &score); 35 tmp.push_back(node(ID, score,loc_num)); 36 } 37 sort(tmp.begin(), tmp.end(), Cmp); 38 vector<node>::iterator it = tmp.begin(); 39 int pre = -1,cnt=0,prer; 40 for (; it != tmp.end(); it++) 41 { 42 if (pre==-1||it->score < pre) it->loc_rank =prer= ++cnt,pre=it->score; 43 else it->loc_rank = prer,cnt++; 44 } 45 vector<node>tans; 46 tans.resize(ans.size() + tmp.size()); 47 merge(ans.begin(), ans.end(), tmp.begin(), tmp.end(), tans.begin(),Cmp); 48 ans.clear(); 49 ans.assign(tans.begin(), tans.end()); 50 loc_num++; 51 } 52 printf("%d ", ans.size()); 53 int prescore = -1, prer, cnt = 0; 54 vector<node>::iterator it = ans.begin(); 55 for (; it != ans.end(); it++) 56 { 57 printf("%s", it->id.c_str()); 58 if (prescore == -1 || it->score < prescore) 59 { 60 prer = ++cnt; 61 prescore = it->score; 62 } 63 else cnt++; 64 printf(" %d", prer); 65 printf(" %d %d ", it->loc_num, it->loc_rank); 66 } 67 return 0; 68 }
1026. Table Tennis
题意：有k个台球桌，其中有m个为VIP桌。该台球室营业时间为9:00:00到21:00:00。有若干pair来该台球室打球。如果台球桌已满人，否则使用编号最小的台球桌。这些人当中有VIP用户。如果在等待时有个台球桌刚好空出来，且为VIP桌，则VIP用户可率先使用，有多个VIP用户时按到达时间的顺序使用。给出该台球室按每个桌子的开始服务时间排序的使用情况（所服务pair的到来时间、开始服务时间、等待时间），并给出每个桌子服务pair数目。
思路：优先队列维护当前可使用的台球桌，对用户的到来时间进行排序。如果到来有桌子为空，则可立马使用；否则入等待队列；等待时若有桌子空出，则若其为VIP桌且等待队伍中有VIP用户，则由该VIP用户先用，否则由队伍的第一个人使用。
注意：对等待时间四舍五入；每个pair的使用时间最多2小时；不在营业时间到来的客户或者营业时间结束停止服务。
1 #include<iostream> 2 #include<vector> 3 #include<cstdio> 4 #include<algorithm> 5 #include<queue> 6 using namespace std; 7 struct node 8 { 9 int ar_time; 10 int pl_time; 11 bool vip; 12 node(int aa = 0, int pp = 0, bool vv = false) :ar_time(aa), pl_time(pp), vip(vv) {} 13 }; 14 vector<node>customers; 15 vector<node>waitVec; 16 int tables[110]; 17 int viptab[110]; 18 struct tab 19 { 20 int endtime; 21 int id; 22 tab(int ee = 0, int ii = 0) :endtime(ee), id(ii) {} 23 }; 24 vector<tab>pq; 25 bool Cmp2(const tab&a, const tab&b) 26 { 27 if (a.endtime == b.endtime) 28 { 29 return a.id > b.id; 30 } 31 else return a.endtime > b.endtime; 32 } 33 bool Cmp(node&a, node&b) 34 { 35 return a.ar_time < b.ar_time; 36 } 37 char ar[10]; 38 int exChange() 39 { 40 return ar[7] - '0' + (ar[6] - '0') * 10 + (ar[4] - '0') * 60 + (ar[3] - '0') * 10 * 60 + (ar[1] - '0') * 3600 + (ar[0] - '0') * 10 * 3600; 41 } 42 void int2str(int time) 43 { 44 ar[0] = '0' + time / 3600 / 10; 45 ar[1] = '0' + time / 3600 % 10; 46 time -= (ar[1] - '0') * 3600 + (ar[0] - '0') * 10 * 3600; 47 ar[3] = '0' + time / 60 / 10; 48 ar[4] = '0' + time / 60 % 10; 49 time -= (ar[4] - '0') * 60 + (ar[3] - '0') * 10 * 60; 50 ar[6] = '0' + time / 10; 51 ar[7] = '0' + time % 10; 52 ar[2] = ar[5] = ':'; 53 ar[8] = ' '; 54 } 55 56 void run(int ttime) 57 { 58 tab ff = pq.front(); 59 while (ff.endtime < ttime&&waitVec.size()) 60 { 61 62 int index = 0; 63 node cur = waitVec[0]; 64 bool isvip_inwait = false; 65 vector<node>::iterator it = waitVec.begin(); 66 for (; it != waitVec.end(); it++) 67 { 68 if (it->vip) 69 { 70 index = it - waitVec.begin(); 71 cur = waitVec[index]; 72 isvip_inwait = true; 73 break; 74 } 75 } 76 if (isvip_inwait) 77 { 78 vector<tab>::iterator itab = pq.begin(); 79 for (; itab != pq.end(); itab++) 80 { 81 if (itab->endtime == ff.endtime) 82 { 83 if (viptab[itab->id]) 84 { 85 int tt = ff.id; 86 ff.id =pq.begin()->id=itab->id; 87 itab->id = tt; 88 break; 89 } 90 } 91 } 92 } 93 if(!viptab[ff.id]) index = 0,cur = waitVec[0]; 94 95 int2str(cur.ar_time); 96 printf("%s", ar); 97 if (ff.endtime >= cur.ar_time)int2str(ff.endtime); 98 int waitTime = ((ff.endtime <= cur.ar_time) ? 0 : ff.endtime - cur.ar_time); 99 printf(" %s %d ", ar, (waitTime + 30) / 60); 100 waitVec.erase(waitVec.begin() + index); 101 pop_heap(pq.begin(), pq.end(), Cmp2); 102 pq.pop_back(); 103 int endTime = ((ff.endtime <= cur.ar_time) ? cur.ar_time + cur.pl_time : ff.endtime + cur.pl_time); 104 pq.push_back(tab(endTime, ff.id)); 105 push_heap(pq.begin(), pq.end(), Cmp2); 106 tables[ff.id]++; 107 ff = pq.front(); 108 } 109 } 110 int main() 111 { 112 int n, Count = 0; 113 int st = 8 * 3600, ed = 21 * 3600; 114 scanf("%d", &n); 115 for (int i = 1; i <= n; i++) 116 { 117 int pl, vip; 118 scanf("%s%d%d", ar, &pl, &vip); 119 if (pl > 120) pl = 120; 120 pl = pl * 60; 121 int artime = exChange(); 122 if (artime >= ed) continue; 123 customers.push_back(node(artime, pl, vip)); 124 Count++; 125 } 126 sort(customers.begin(), customers.end(), Cmp); 127 int k, m; 128 scanf("%d%d", &k, &m); 129 for (int i = 1; i <= m; i++) 130 { 131 int id; 132 scanf("%d", &id); 133 viptab[id] = 1; 134 } 135 136 137 for (int i = 1; i <= k; i++) 138 { 139 pq.push_back(tab(st, i)); 140 } 141 make_heap(pq.begin(), pq.end(), Cmp2); 142 for (int i = 0; i < Count;) 143 { 144 tab ff = pq.front(); 145 if (waitVec.size() == 0 || customers[i].ar_time <= ff.endtime) 146 { 147 waitVec.push_back(customers[i]); 148 i++; 149 } 150 else 151 { 152 run(customers[i].ar_time); 153 } 154 } 155 run(ed); 156 for (int i = 1; i <= k; i++) 157 { 158 if (i> 1) printf(" "); 159 printf("%d", tables[i]); 160 } 161 printf(" "); 162 return 0; 163 }
1027. Colors in Mars
题意：把给出的3个十进制数转为3个2为13进制数。
思路：对13整除和取余。
1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 char color[10]; 5 int main() 6 { 7 for (int i = 0; i < 3; i++) 8 { 9 int v; 10 scanf("%d", &v); 11 color[i * 2] = ((v / 13 <= 9) ? '0' + v / 13 : 'A' + v / 13 - 10); 12 color[i * 2 + 1] = ((v % 13 <= 9) ? '0' + v % 13 : 'A' + v % 13 - 10); 13 } 14 printf("#%s ", color); 15 return 0; 16 }