A:門牌製做
題面：
思路：
暴力遍歷，1-2020的數字，而後判斷每一位，計數。c++
代碼：
/* * @Date: 2020-10-17 21:49:55 * @LastEditTime: 2020-10-17 21:58:58 * @Author's blog: blog.nuoyanli.com * @Description: Plum blossom from the bchter cold! */ #include <bits/stdc++.h> using namespace std; int main() { int ans = 0; for (int i = 1; i <= 2020; i++) { int j = i; while (j) { if (j % 10 == 2){ ans++; } j /= 10; } } cout << ans << endl; return 0; }
答案：
624spa
B:既約分數
思路：
遍歷1-2020之間的任意點對，判斷gcd是否爲1，計數。code
代碼：
/* * @Date: 2020-10-17 22:02:12 * @LastEditTime: 2020-10-17 22:02:24 * @Author's blog: blog.nuoyanli.com * @Description: Plum blossom from the bchter cold! */ #include <bits/stdc++.h> using namespace std; int main() { int ans = 0; for (int i = 1; i <= 2020; i++) { for (int j = 1; j <= 2020; j++) { if (__gcd(i, j) == 1){ ans++; } } } cout << ans << endl; return 0; }
答案:
2481215blog
C:蛇形填數
題面：
思路：
模擬整個過程，判斷一下奇偶位置的特殊狀況。圖片
代碼：
/* * @Date: 2020-10-17 22:05:26 * @LastEditTime: 2020-10-17 22:06:55 * @Author's blog: blog.nuoyanli.com * @Description: Plum blossom from the bchter cold! */ #include <bits/stdc++.h> using namespace std; int mmp[50][50], id; int main() { for (int i = 1; i <= 40; i++) { if (i & 1) { int idx = i, idy = 1; for (int j = 0; j < i; j++){ mmp[idx - j][idy + j] = ++id; } } else { int idx = 1, idy = i; for (int j = 0; j < i; j++){ mmp[idx + j][idy - j] = ++id; } } } cout << mmp[20][20] << endl; return 0; }
答案：
761ip
D:跑步鍛鍊
題面：
思路：
直接模擬日期加法，遞增到2020年10月1日爲止，計數（注意第一天也是2哦），注意判斷閏年，處理好年月日。it
代碼：
/* * @Date: 2020-10-17 22:09:57 * @LastEditTime: 2020-10-17 22:11:55 * @Author's blog: blog.nuoyanli.com * @Description: Plum blossom from the bchter cold! */ #include <bits/stdc++.h> using namespace std; int M[13] = { 0, 31, 0, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; int main() { int y = 2000, m = 1, d = 1, w = 6, ans = 2; while (y != 2020 || m != 10 || d != 1) { if (y % 400 == 0 || (y % 4 == 0 && y % 100)) { M[2] = 29; } else { M[2] = 28; } d++; w = (w + 1) % 7; if (d > M[m]) { d = 1; m++; if (m > 12) { m = 1; y++; } } if (d == 1 || w == 1) { ans++; } ans++; } cout << ans << endl; return 0; }
答案：
8879io
E:七段碼
題面：
思路：
暴力枚舉 2 7 = 128 2^7=128 27=128次方的全部七段碼，而後DFS判斷聯通塊。（不愧是暴力杯 ）ast
代碼：
建圖代碼:
/* * @Date: 2020-10-17 21:20:30 * @LastEditTime: 2020-10-17 22:15:57 * @Author's blog: blog.nuoyanli.com * @Description: Plum blossom from the bchter cold! */ #include <bits/stdc++.h> using namespace std; int main() { int ans = 0; freopen("in.txt", "w", stdout); for (int a = 0; a <= 1; a++) { for (int b = 0; b <= 1; b++) { for (int c = 0; c <= 1; c++) { for (int d = 0; d <= 1; d++) { for (int e = 0; e <= 1; e++) { for (int f = 0; f <= 1; f++) { for (int g = 0; g <= 1; g++) { int mmp[6][4] = { 0}; if (a) { for (int i = 1; i <= 3; i++) { mmp[1][i] = 1; } } if (b) { for (int i = 1; i <= 3; i++) { mmp[i][3] = 1; } } if (c) { for (int i = 3; i <= 5; i++) { mmp[i][3] = 1; } } if (d) { for (int i = 1; i <= 3; i++) { mmp[5][i] = 1; } } if (e) { for (int i = 3; i <= 5; i++) { mmp[i][1] = 1; } } if (f) { for (int i = 1; i <= 3; i++) { mmp[i][1] = 1; } } if (g) { for (int i = 1; i <= 3; i++) { mmp[3][i] = 1; } } printf("5 3\n"); for (int i = 1; i <= 5; i++) { for (int j = 1; j <= 3; j++) { printf("%d", mmp[i][j]); } puts(""); } puts(""); } } } } } } } return 0; }
DFS求聯通塊代碼
/* * @Date: 2020-10-17 21:20:03 * @LastEditTime: 2020-10-17 21:27:00 * @Author's blog: nuoyanli * @Description: Plum blossom from the bchter cold! */ #include <bits/stdc++.h> using namespace std; const int N = 10; char a[N][N]; int n, m, vis[N][N]; void dfs(int r, int c, int f) { if (r < 0 || r >= m || c < 0 || c >= n) { return; } if (vis[r][c] > 0 || a[r][c] != '1') { return; } vis[r][c] = f; for (int i = -1; i <= 1; i++){ for (int j = -1; j <= 1; j++){ if (i != 0 || j != 0){ dfs(r + i, c + j, f); } } } } int main() { int sum = 0; freopen("in.txt", "r", stdin); while (~scanf("%d%d", &m, &n)) { for (int i = 0; i < m; i++){ scanf("%s", a[i]); } memset(vis, 0, sizeof(vis)); int cnt = 0; for (int i = 0; i < m; i++){ for (int j = 0; j < n; j++){ if (vis[i][j] == 0 && a[i][j] == '1'){ dfs(i, j, ++cnt); } } } if (cnt == 1){ sum += 1; } } printf("%d\n", sum); return 0; }
F:成績統計
題面：
思路：
直接計算，而後注意四捨五入。class
代碼：
/* * @Date: 2020-10-17 22:18:36 * @LastEditTime: 2020-10-17 22:21:48 * @Author's blog: blog.nuoyanli.com * @Description: Plum blossom from the bchter cold! */ #include <bits/stdc++.h> using namespace std; int main() { int n, jg = 0, yx = 0; scanf("%d", &n); while (n--) { int x; scanf("%d", &x); if (x >= 60) { jg++; } if (x >= 85) { yx++; } } double Jg = jg * 1.0 / n * 1.0; double Yx = yx * 1.0 / n * 1.0; int ans_jg = int(Jg * 1000 + 5) / 10; int ans_yx = int(Yx * 1000 + 5) / 10; printf("%d%%\n%d%%\n", ans_jg, ans_yx); return 0; }
H:子串分值和
題面：
思路：
算單字母的貢獻，而後記錄一下上一個位置。
代碼：
/* * @Date: 2020-10-18 13:13:49 * @LastEditTime: 2020-10-18 15:26:45 * @Author's blog: blog.nuoyanli.com * @Description: Plum blossom from the bchter cold! */ #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 1e6 + 10; char s[N]; ll vis[40]; int main() { scanf("%s", s + 1); int n = strlen(s + 1); ll ans = 0; for (int i = 1; i <= n; i++) { ans += (i - vis[s[i] - 'a']) * (n - i + 1); vis[s[i] - 'a'] = i; } cout << ans << endl; return 0; }