/*读了老半天才把题读懂，读懂了题输出格式没注意，结果re了两次。 题意：先给一串数字S，然后每次给出对应相同数目的的一串数字Si，然后优先统计Si和S对应位相同并且相等的个数L，在统计不在对应位上但相等的的个数R. 当Si全0时，表示结束。 每个数只能用一次。 例如：有 S 1 3 5 5 S1 1 1 2 3 S2 4 3 3 5 Si . . . . 0 0 0 0 对于S1:则第一步只有第一个对应相等，所以L = 1 就还剩下3 5 5 1 2 3，在统计R，下面只有3和上面的3相同，所以R = 1; 对于S2:L = 2,第二个位置和第四个位置对应相等。 就还剩下1 5 4 3,所以R = 0;依次类推。 */ #include <iostream> #include <fstream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <climits> #include <cstring> #include <string> #define N 1005 using namespace std; int n; int a[N], b[N], c[N]; int main() { int i, j, t = 1, l, r; while (cin>>n, n) { cout<<"Game "<<t++<<':'<<endl; for (i = 0; i < n; i++) cin>>a[i]; while (true) { for (i = 0; i < n; i++) c[i] = a[i]; l = r = 0; for (i = 0; i < n; i++) { cin>>b[i]; if (b[i] == c[i])/**< 统计对应位相等l的个数 */ { l++; c[i] = 0;/**< 置相等位0,统计r时不影响 */ b[i] = -1; } } if (!b[0]) break; for (i = 0; i < n; i++) { if (b[i] == -1) continue; for (j = 0; j < n; j++) { if (b[i] == c[j])/**< 统计r的个数 */ { r++; c[j] = 0; break; } } } cout<<" "<<'('<<l<<','<<r<<')'<<endl; } } return 0; }
1 /*题意:统计每个国家有多少个人被Giovanni喜欢。 2 3 并按字典序输出。 4 */ 5 6 #include <iostream> 7 #include <string> 8 #include <cstring> 9 #include <cstdio> 10 #include <cstdlib> 11 #include <fstream> 12 #include <climits> 13 #define N 2005 14 using namespace std; 15 struct node 16 { 17 string c; 18 int sum; 19 node() 20 { 21 c.clear(); 22 sum = 0; 23 } 24 bool operator >= (const node &b) 25 { 26 return c >= b.c; 27 } 28 friend istream& operator >> (istream &in, node &res) 29 { 30 in>>res.c; 31 return in; 32 } 33 friend ostream &operator << (ostream &out, const node &res) 34 { 35 out<<res.c<<' '<<res.sum<<endl; 36 return out; 37 } 38 bool operator == (const node &b) 39 { 40 return c == b.c; 41 } 42 43 bool operator != (const node &b) 44 { 45 return c != b.c; 46 } 47 }; 48 49 struct li 50 { 51 node s[N]; 52 int p; 53 li() 54 { 55 p = 0; 56 } 57 friend ostream &operator << (ostream &out, const li &res) 58 { 59 for (int i = 0; i < res.p; i++) 60 out<<res.s[i]; 61 } 62 void insert(const node &res) 63 { 64 if (p) 65 { 66 int l = 0, r = p - 1, mid; 67 while (l <= r)/**< 二分查找第一个比res大或等于的位置 */ 68 { 69 mid = (l + r) >> 1; 70 if (s[mid] >= res)/**< 如果大于等于，在左区间位置 */ 71 r = mid - 1; 72 else/**< 在右区间 */ 73 l = mid + 1; 74 } 75 if (s[l] != res)/**< 不存在，插入 */ 76 { 77 for (r = p++; r > l; r--) 78 s[r] = s[r - 1]; 79 s[l] = res; 80 } 81 s[l].sum++; 82 return; 83 } 84 s[p] = res; 85 s[p++].sum++; 86 } 87 }; 88 li a; 89 int main() 90 { 91 int n, i; 92 cin>>n; 93 string tt; 94 node t; 95 for (i = 0; i < n; i++) 96 { 97 cin>>t; 98 getline(cin, tt); 99 a.insert(t); 100 } 101 cout<<a; 102 return 0; 103 }
/**< 题意:给一组数据无序，查询给定数在这组数出现的第一个位置(在有序中) */ #include <iostream> #include <string> #include <cstring> #include <cstdio> #include <cstdlib> #include <fstream> #include <climits> #define N 10005 using namespace std; int n, q; int sum[N], con[N]; int main() { int t = 1, i, m; while (cin>>n>>q, n || q) { memset(con, 0, sizeof(con)); cout<<"CASE# "<<t++<<':'<<endl; for (i = 0; i < n; i++) { cin>>m; con[m]++;/**< 计数排序,每个数不超过10000 */ } sum[0] = 0; sum[1] = con[0]; for (i = 2; i < N; i++) sum[i] = sum[i - 1] + con[i - 1]; for (i = 0; i < q; i++) { cin>>m; if (con[m]) cout<<m<<" found at "<<sum[m] + 1<<endl; else cout<<m<<" not found"<<endl; } } return 0; }
/**< 题意:(又是读了半天也没读懂，好纠结的英语) 在三维空间给一些点，计算每个点与其他点的距离，但只取最短距离，并统计最短距离在[0-10)的个数 比如某个点的最短距离在[1,2)之间，则对1这个位置的个数加1 */ #include <iostream> #include <string> #include <cstring> #include <cstdio> #include <cstdlib> #include <fstream> #include <climits> #include <cmath> #include <algorithm> #include<iomanip> #define N 5005 using namespace std; struct node { int x, y, z; /* bool operator > (const node &b) { if (x == b.x) { if (y == b.y) return z > b.z; return y > b.y; } return x > b.x; }*/ }a[N]; int con[11]; int dis(const node &a, const node &b) { int x = a.x - b.x, y = a.y - b.y, z = a.z - b.z; return int(sqrt(x * x + y * y + z * z)); } int cmp(const void *a, const void *b)/**< */ { return ((node *)a)->x > ((node *)b)->x ? 1 : -1; } int main() { int n = 0, i, j, tm, t; while (cin>>a[n].x>>a[n].y>>a[n].z, (a[n].x || a[n].y || a[n].z)) { n++; } //qsort(a, n, sizeof(a[0]), cmp); memset(con, 0, sizeof(con)); for (i = 0; i < n; i++) { tm = 10; for (j = 0; j < n; j++) { if (i != j) { t = dis(a[i], a[j]); if (t < tm) tm = t; } } con[tm]++; } for (i = 0; i < 10; i++) cout<<setw(4)<<con[i]; cout<<endl; return 0; }
/**< 题目啰嗦了半天，就是火车进行从小到大重排序(冒泡排序)交换的次数 */ #include <iostream> #include <algorithm> using namespace std; #define N 55 int a[N]; int main() { int n, l, i, j, s; cin>>n; while (n--) { cin>>l; for (i = 0; i < l; i++) cin>>a[i]; s = 0; for (i = 0; i < l; i++) for (j = l - 1; j > i; j--) { if (a[j] < a[j - 1]) { swap(a[j], a[j - 1]); s++; } } cout<<"Optimal train swapping takes "<<s<<" swaps."<<endl; } return 0; }
/**< 翻转从小到大排序*/ #include <iostream> #include <algorithm> #include <cstdio> using namespace std; #define N 35 int a[N]; int main() { int i, j, n = 0, p, mm, t; char ch = 1; while (ch != EOF && cin>>a[n]) { ch = cin.get(); if (ch == ' ') { n++; continue; } t = ++n; cout<<a[0]; for (i = 1; i < n; i++) cout<<' '<<a[i]; cout<<endl; for (i = 0 ; i < n; n--) { mm = a[i]; p = 0; for (j = 1; j < n; j++) { if (mm < a[j]) { mm = a[j]; p = j; } } if (p != n - 1) { if (p) { reverse(a, a + p + 1); reverse(a, a + n); cout<<t - p<<' '<<t - n + 1<<' '; } else { reverse(a, a + n); cout<<t - n + 1<<' '; } } } cout<<0<<endl; } return 0; }
/**< 给一些单词，去掉相同字母且个数一样不区分大小写，再给剩下的单词进行排序输出 */ #include <iostream> #include <fstream> #include <string> #include <cstring> #include <cstdio> #include <cstdlib> #include <algorithm> #define N 1005 using namespace std; struct node { string str; bool flag; int tag[26]; bool operator == (const node &b) { for (int i = 0; i < 26; i++) if (tag[i] != b.tag[i]) return false; return true; } bool operator > (const node &b) { return str > b.str; } node(const string &s) { str = s; flag = false; memset(tag, 0, sizeof(tag)); for (int i = 0; s[i]; i++) tag[(s[i] | 32) - 'a']++; } node() { flag = false; memset(tag, 0, sizeof(tag)); } }s[N]; int p = 0; void insert(const string &v) { if (!p) { s[p++] = v; return; } for (int i = 0; i < p; i++) if (s[i] == v) { s[i].flag = true; return; } int l = 0, r = p - 1, mid; while (l <= r) { mid = (l + r) >> 1; if (s[mid] > v) r = mid - 1; else l = mid + 1; } for (r = p++; r > l; r--) s[r] = s[r - 1]; s[l] = v; } string t; int main() { int i; char ch; cin.get(ch); while (ch != '#') { t.clear(); while (ch == ' ' || ch == ' ') cin.get(ch); if (ch == '#') break; while (true) { t += ch; cin.get(ch); if (ch == ' ' || ch == ' ') break; } insert(t); } for (i = 0; i < p; i++) if (!s[i].flag) cout<<s[i].str<<endl; return 0; }
/**< 对输入的字符串进行排序，按格式输出，最后一列的宽度为所有字符串最长宽度，其余列的宽度为最长宽度加2. 要求输出所用的行最少 */ #include <iostream> #include <fstream> #include <string> #include <cstring> #include <cstdio> #include <cstdlib> #include <algorithm> #include <iomanip> using namespace std; #define N 105 string s[N]; int main() { int n, i, mm, len, j, t, k, kk; while (cin>>n) { mm = 0; for (i = 0; i < n; i++) { cin>>s[i]; len = s[i].length(); if (mm < len) mm = len; } sort(s, s + n); len = (60 - mm) / (mm + 2) + 1; for (i = 0; i < 60; i++) cout<<'-'; cout<<endl; cout<<left; t = n / len;/**< 要输出t行len列 */ if (n % len) t++; for (i = 0; i < t; i++) { kk = mm + 2; for (j = i, k = 1; j < n; j += t, k++) { if (k == len)/**< 最后一列输出宽度为mm */ kk = mm; cout<<setw(kk)<<s[j]; } cout<<endl; } } return 0; }
#include <iostream> #include <fstream> #include <string> #include <cstring> #include <cstdio> #include <cstdlib> #include <algorithm> #include <iomanip> #include <set> using namespace std; struct snode { string str; int x, y;/**< 字符串位置 */ bool operator < (const snode &b)const { if (str == b.str) { if (x == b.x) return y < b.y; return x < b.x; } return str < b.str; } snode() { str.clear(); x = y = 0; } snode(string v, int xx, int yy) { str = v; x = xx; y = yy; } friend ostream & operator << (ostream &out, const snode &v) { out<<v.str<<' '<<v.x<<' '<<v.y; return out; } }; set<string> ks; multiset<snode> res; string ts[205]; int main() { int i, j, p; string tmp; ks.clear(); res.clear(); while (cin>>tmp, tmp != "::") { ks.insert(tmp); } i = 0; cin.get(); while (getline(cin, ts[i])) { j = 0; while (true) { p = j; tmp.clear(); while (ts[i][j] && ts[i][j] != ' ') { ts[i][j] |= 32; tmp += ts[i][j++]; } if (ks.find(tmp) == ks.end())/**< 不存在 */ res.insert(snode(tmp, i, p)); if (!ts[i][j]) break; j++; } i++; } for (multiset<snode>::iterator it = res.begin(); it != res.end(); it++) { for (i = 0; i < it->y; i++) cout<<ts[it->x][i]; for (; ts[it->x][i] != ' ' && ts[it->x][i]; i++)/**< 大写 */ cout<<char(ts[it->x][i] & 95); for (; ts[it->x][i]; i++) cout<<ts[it->x][i]; //cout<<*it; cout<<endl; } return 0; }
#include <iostream> #include <cstdio> #include <fstream> #include <cstring> #include <string> #include <cctype> #include <iomanip> #include <cstdlib> #include <algorithm> #include <set> #include <map> using namespace std; struct snode { snode() { score = in = out = win = con = tie = lose = 0; name.clear(); _name.clear(); } snode(string v) { score = in = out = win = con = tie = lose = 0; name = v; _name.clear(); for (int i = 0; v[i]; i++) { if (isalnum(v[i])) _name += (v[i] | 32); else _name += v[i]; } } string name; string _name; int score;/**< 总得分 */ int in;/**< 进球 */ int out;/**< 对方进球 */ int win;/**< 赢得场数 */ int con;/**< 参加比赛次数 */ int tie;/**< 平局次数 */ int lose;/**< 输的次数 */ bool operator < (const snode &b)const { if (score == b.score) { if (win == b.win) { int ta = in - out, tb = b.in - b.out; if (ta == tb) { if (in == b.in) { if (con == b.con) { return _name < b._name; } return con < b.con; } return in > b.in; } return ta > tb; } return win > b.win; } return score > b.score; } friend istream &operator >> (istream &in, snode &v) { string tmp; getline(in, tmp); v = tmp; return in; } friend ostream &operator << (ostream &out, const snode &v) { cout<<v.name<<' '<<v.score<<"p, "<<v.con<<"g ("<<v.win<<'-'<<v.tie<<'-'<<v.lose<<"), "; cout<<v.in - v.out<<"gd ("<<v.in<<'-'<<v.out<<')'; return out; } bool operator == (const snode &v) { return name == v.name; } }a[35]; int n, t, g; int find_str(string v) { for (int i = 0; i < t; i++) if (a[i] == v) return i; return -1; } void fun(string v) { int i = -1; while (v[++i] != '#'); string s1(v.begin(), v.begin() + i); /**< 数字处 */ int t1 = v[++i] - '0'; if (isdigit(v[++i]))/**< 第二个是不是数字 */ t1 = 10 * t1 + (v[i++] - '0'); int t2 = v[++i] - '0'; if (isdigit(v[++i]))/**< 第二个是不是数字 */ t2 = 10 * t2 + (v[i++] - '0'); string s2(v.begin() + i + 1, v.end()); int p1 = find_str(s1), p2 = find_str(s2); //cout<<p1<<p2<<endl; if (t1 > t2)/**< 1赢 */ { a[p1].win++; a[p2].lose++; a[p1].score += 3; } else if (t1 == t2)/**< 平 */ { a[p1].tie++; a[p2].tie++; a[p1].score++; a[p2].score++; } else/**< 2赢 */ { a[p2].win++; a[p1].lose++; a[p2].score += 3; } a[p1].con++; a[p2].con++; a[p1].in += t1; a[p2].in += t2; a[p1].out += t2; a[p2].out += t1; } int main() { int i, j; cin>>n; cin.get(); string st; while (n--) { getline(cin, st); cout<<st<<endl; cin>>t; cin.get(); for (i = 0; i < t; i++) cin>>a[i]; cin>>g; cin.get(); for (i = 0; i < g; i++) { getline(cin, st); fun(st); } sort(a, a + t); for (i = 0; i < t; i++) { cout<<i + 1<<") "<<a[i]<<endl; } if (n) cout<<endl; } return 0; }
#include <iostream> #include <cstdio> #include <fstream> #include <cstring> #include <string> #include <cctype> #include <iomanip> #include <cstdlib> #include <algorithm> #include <set> #include <map> using namespace std; int ls[] = { 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 0, 7, 7, 8, 8, 8, 9, 9, 9, 0 }; char get(char ch) { return ls[ch - 'A'] + '0'; } map<string, int> mp; int main() { int t, n, i, tag; cin>>t; char ch; string tmp; while (t--) { mp.clear(); cin>>n; cin.get(); for (i = 0; i < n; i++) { tmp.clear(); while (ch = cin.get(), ch != ' ')/**< 对输入进行处理换成全数字的字符串 */ { if (isdigit(ch)) { tmp += ch; continue; } if (isalpha(ch)) { tmp += get(ch); continue; } } mp[tmp]++;/**< 对应字符串个数加1 */ } tag = true; for (map<string, int>::iterator it = mp.begin(); it != mp.end(); ++it) { if (it->second > 1) { tmp = it->first; tmp.insert(tmp.begin() + 3, '-'); cout<<tmp<<' '<<it->second<<endl; tag = false; } } if (tag) cout<<"No duplicates."<<endl; if (t) cout<<endl; } return 0; }
#include <iostream> #include <fstream> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <iomanip> #include <cctype> #include <climits> #include <algorithm> #include <map> #include <set> using namespace std; map<char, int> mp1, mp2; char s1[] = { 'J', 'S', 'B', 'K', 'T', 'C', 'L', 'D', 'M', 'V', 'N', 'W', 'F', 'X', 'G', 'P', 'Y', 'H', 'Q', 'Z', 'R' }; char s2[] = { 'A' , 'U', 'E', 'O', 'I' }; int n1, n2, p1, p2; void fun1() { while (n1 > 0) { p1++; if (n1 >= 5) mp1[s1[p1]] = 5; else mp1[s1[p1]] = n1; n1 -= 5; } } void fun2() { while (n2 > 0) { p2++; if (n2 >= 21) mp2[s2[p2]] = 21; else mp2[s2[p2]] = n2; n2 -= 21; } } void print() { int i; string s1, s2; s1.clear(); s2.clear(); for (map<char, int>::iterator it = mp1.begin(); it != mp1.end(); ++it) { for (i = it->second; i > 0; i--) s1 += it->first; } for (map<char, int>::iterator it = mp2.begin(); it != mp2.end(); ++it) { for (i = it->second; i > 0; i--) s2 += it->first; } i = 0; while (s1[i] && s2[i]) { cout<<s2[i]<<s1[i]; i++; } if (s2[i]) cout<<s2[i]; cout<<endl; } int main() { int m, n, t = 1; cin>>m; while (m--) { cout<<"Case "<<t++<<": "; cin>>n; p1 = p2 = -1; n1 = n / 2; n2 = n1 + (n & 1); mp1.clear(); mp2.clear(); fun1(); fun2(); print(); } return 0; }