标签：open -- print break eve def pair 连续 main
A：
水。
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const ll INFLL = 0x3f3f3f3f3f3f3f3f; 8 9 ll n, m, h, w; 10 11 ll solve() 12 { 13 ll cnt = 0; 14 ll tn = n, tm = m; 15 while(tn > h) 16 { 17 tn = (tn + 1) >> 1; 18 cnt++; 19 } 20 while(tm > w) 21 { 22 tm = (tm + 1) >> 1; 23 cnt++; 24 } 25 return cnt; 26 } 27 28 int main() 29 { 30 while(~scanf("%lld %lld %lld %lld", &n, &m, &h, &w)) 31 { 32 ll ans = INFLL; 33 ans = min(ans, solve()); 34 swap(h, w); 35 ans = min(ans, solve()); 36 printf("%lld\n", ans); 37 } 38 return 0; 39 }
B：留坑。
C:留坑。
D：
题意：给出一棵树中每个点的度数，构造这一棵树
思路：将点按度数从大到小排序，每个点会有一个父亲，最多连出n - 1个点。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 200010 5 #define pii pair <int, int> 6 int n, d[N], pos[N], fa[N]; 7 8 bool cmp (int a, int b) { return d[a] > d[b]; } 9 10 int main() 11 { 12 while (scanf("%d", &n) != EOF) 13 { 14 for (int i = 1; i <= n; ++i) scanf("%d", d + i), pos[i] = i; 15 sort(pos + 1, pos + 1 + n, cmp); 16 for (int i = 2; i <= n; ++i) --d[pos[i]]; 17 int id = 1; 18 for (int i = 1; i <= n; ++i) 19 { 20 while (d[pos[i]]--) 21 fa[pos[++id]] = pos[i]; 22 } 23 for (int i = 1; i <= n; ++i) if (i != pos[1]) 24 printf("%d %d\n", fa[i], i); 25 } 26 return 0; 27 }
E：
水。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 1000010 6 char s[N]; 7 8 int main() 9 { 10 while (scanf("%s", s + 1) != EOF) 11 { 12 int cnt = 0; 13 for (int i = 1, len = strlen(s + 1); i <= len; ++i) cnt += (s[i] == ‘A‘); 14 ll res = 0; 15 for (ll i = 1; ; ++i) 16 { 17 if (i * (i + 1) / 2 > cnt) break; 18 res = i; 19 } 20 printf("%lld\n", res); 21 } 22 return 0; 23 }
F：留坑。
G：
题意：给出一个n 要构造一个长度为n的序列，满足题目要求
思路：
如果是奇数
那么存在 起始点3 有 $2 \cdot 5 \cdot 6 = 3 \cdot 4 \cdot 5$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int n; 5 6 int main() 7 { 8 while (scanf("%d", &n) != EOF) 9 { 10 if (n == 1) puts("NO"); 11 else if (n & 1) 12 { 13 puts("YES"); 14 puts("3"); 15 printf("-++"); 16 for (int i = 4; i <= n; ++i) printf("%c", "+-"[i & 1]); 17 puts(""); 18 } 19 else 20 { 21 puts("YES"); 22 puts("1"); 23 for (int i = 1; i <= n; ++i) printf("%c", "-+"[i & 1]); 24 puts(""); 25 } 26 } 27 return 0; 28 }
H：
题意：给出0-9每个数字的个数，构造一个最大的数，使得任意连续三位连起来都是3的倍数。数字可以不用完
思路：先将所有数字模3
就是三种数字 0， 1， 2
再考虑 排列
显然 三个一组三个一组，三个中012至少出现一次，即012的全排列，一共6种，
再考虑 全放0, 1, 2 又是三种
再考虑 只有一位或者两位，可以任意摆放
一共十种情况，去掉前导0之后取最大即可。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 int cnt[15], tmp_cnt[15], cntt[3], tmp_cntt[3], tar[3], Max[3], tmp_Max[3]; 6 char ans[10][N]; 7 int pos; 8 9 void g() 10 { 11 for (int i = 0; i < 15; ++i) tmp_cnt[i] = cnt[i]; 12 for (int i = 0; i < 3; ++i) 13 { 14 tmp_cntt[i] = cntt[i]; 15 tmp_Max[i] = Max[i]; 16 } 17 } 18 19 void f() 20 { 21 for (int i = 0; i < 15; ++i) cnt[i] = tmp_cnt[i]; 22 for (int i = 0; i < 3; ++i) 23 { 24 cntt[i] = tmp_cntt[i]; 25 Max[i] = tmp_Max[i]; 26 } 27 } 28 29 void solve(int ide, int *tar) 30 { 31 pos = 0; 32 while (1) 33 { 34 for (int i = 0; i < 3; ++i) 35 { 36 int id = tar[i]; 37 if (!cntt[id]) 38 return; 39 if (cnt[Max[id]] == 0) 40 { 41 Max[id] = -1; 42 for (int i = 9; i >= 0; --i) if (i % 3 == id && cnt[i]) 43 { 44 Max[id] = i; 45 break; 46 } 47 if (Max[id] == -1) return; 48 } 49 ans[ide][++pos] = Max[id] + ‘0‘; 50 --cntt[id]; 51 --cnt[Max[id]]; 52 } 53 } 54 } 55 56 void work(int id) 57 { 58 pos = 0; 59 while (cntt[id]) 60 { 61 if (!cnt[Max[id]]) 62 { 63 Max[id] = -1; 64 for (int i = 9; i >= 0; --i) if (i % 3 == id && cnt[i]) 65 { 66 Max[id] = i; 67 break; 68 } 69 if (Max[id] == -1) return; 70 } 71 ans[id + 6][++pos] = Max[id] + ‘0‘; 72 --cnt[Max[id]]; 73 --cntt[id]; 74 } 75 } 76 77 bool cmp(char *a, char *b) 78 { 79 int lena = strlen(a + 1), lenb = strlen(b + 1); 80 if (lena != lenb) return lena < lenb; 81 for (int i = 1; i <= lena; ++i) if (a[i] != b[i]) 82 return a[i] < b[i]; 83 return 1; 84 } 85 86 void cle(int id) 87 { 88 int len = strlen(ans[id] + 1); 89 pos = len + 1; 90 for (int i = 1, len = strlen(ans[id] + 1); i <= len; ++i) if (ans[id][i] != ‘0‘) 91 { 92 pos = i; 93 break; 94 } 95 if (pos == len + 1) 96 { 97 ans[id][1] = ‘0‘; 98 ans[id][2] = 0; 99 } 100 else 101 { 102 for (int i = 1; i + pos - 1 <= len; ++i) 103 ans[id][i] = ans[id][i + pos - 1]; 104 ans[id][len - pos + 2] = 0; 105 } 106 } 107 108 int main() 109 { 110 while (scanf("%d", cnt) != EOF) 111 { 112 memset(cntt, 0, sizeof cntt); 113 memset(Max, -1, sizeof Max); 114 memset(ans, 0, sizeof ans); 115 for (int i = 1; i <= 9; ++i) scanf("%d", cnt + i); 116 for (int i = 0; i < 10; ++i) if (cnt[i]) 117 { 118 cntt[i % 3] += cnt[i]; 119 Max[i % 3] = i; 120 } 121 int res = 0; 122 for (int i = 9; i >= 0 && res < 10; --i) if (cnt[i]) 123 { 124 res = res * 10 + i; 125 if (res < 10 && cnt[i] > 1) 126 res = res * 10 + i; 127 } 128 g(); 129 pos = 0; 130 while (res) 131 { 132 ans[9][++pos] = res % 10 + ‘0‘; 133 res /= 10; 134 } 135 ans[9][pos + 1] = 0; 136 reverse(ans[9] + 1, ans[9] + pos + 1); 137 tar[0] = 0, tar[1] = 1, tar[2] = 2; 138 int ide = 0; 139 do 140 { 141 f(); 142 solve(ide, tar); ans[ide][pos + 1] = 0; 143 ++ide; 144 } while (next_permutation(tar, tar + 3)); 145 for (int i = 0; i < 3; ++i) 146 { 147 f(); pos = 0; work(i); 148 ans[i + 6][pos + 1] = 0; 149 } 150 for (int i = 0; i < 10; ++i) cle(i); 151 pos = 0; 152 for (int i = 1; i < 10; ++i) if (cmp(ans[pos], ans[i])) 153 pos = i; 154 printf("%s\n", ans[pos] + 1); 155 } 156 return 0; 157 }
I：留坑。
J：留坑。
k：留坑。
L：水。
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 1010 6 int n, k, x, cnt; 7 char s[15]; 8 ll Bit[15]; 9 10 ll f(char *s) 11 { 12 ll res = 0; 13 for (int i = 1, len = strlen(s + 1); i <= len; ++i) 14 res = res * 10 + s[i] - ‘0‘; 15 return res; 16 } 17 18 int main() 19 { 20 Bit[1] = 1; 21 for (int i = 2; i <= 10; ++i) Bit[i] = Bit[i - 1] * 10; 22 while (scanf("%d", &n) != EOF) 23 { 24 cnt = 0; ll Min = 0; 25 for (int i = 1; i <= n; ++i) 26 { 27 scanf("%s", s + 1); 28 cnt = max(cnt, (int)strlen(s + 1)); 29 Min = max(Min, f(s)); 30 } 31 Min = max((ll)n, Min); 32 Min = max(Min, Bit[cnt]); 33 printf("%lld\n", Min); 34 for (int i = 1; i <= cnt; ++i) putchar(‘9‘); putchar(‘\n‘); 35 } 36 return 0; 37 }
标签：open -- print break eve def pair 连续 main
原文地址：https://www.cnblogs.com/Dup4/p/9942889.html
