試題A ：門牌制作

// 624 #include <iostream> using namespace std;int solve(int x) {int res = 0;while (x){if (x % 10 == 2) res ++ ;x /= 10;}return res; }int main() {int res = 0;for (int i = 1; i <= 2020; i ++ ){res += solve(i);}cout << res; }

試題B ：既約分數

// 2481215 #include <iostream> using namespace std;int gcd(int a, int b) {return b ? gcd(b, a % b) : a; }int main() {int res = 0;for (int fenzi = 1; fenzi <= 2020; fenzi ++ )for (int fenmu = 1; fenmu <= 2020; fenmu ++ ){if (gcd(fenzi, fenmu) == 1)res ++ ;}cout << res; }

試題C ：蛇形填數

// 761 #include <iostream> using namespace std;int cnt = 1; int g[100][100];int main() {int x = 1, y = 1;while (cnt <= 3000){// 1g[x][y] = cnt ++ ;// 2y ++ ;for ( ; y >= 1; x ++ , y -- )g[x][y] = cnt ++ ;x -- , y ++ ;// 3x ++ ;for ( ; x >= 2; x -- , y ++ )g[x][y] = cnt ++ ;}cout << g[20][20]; }

試題D ：跑步鍛煉

// 8879 #include <iostream> using namespace std;int months[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};int get_day(int year, int month) {if (month != 2) return months[month];if ((year % 400 == 0) || (year % 4 == 0 && year % 100 != 0)) return 29;return 28; }int main() {int sum = 0, res = 0;for (int i = 2000; i <= 2019; i ++ ){for (int j = 1; j <= 12; j ++ ){for (int k = 1; k <= get_day(i, j); k ++ ){int weekend = (sum % 7 + 5) % 7;if (k == 1 || weekend == 0) res += 2;else res ++ ;sum ++ ;}}}for (int i = 1; i <= 9; i ++ ){for (int j = 1; j <= get_day(2020, i); j ++ ){int weekend = (sum % 7 + 5) % 7;if (j == 1 || weekend == 0) res += 2;else res ++ ;sum ++ ;}}res += 2;cout << res; }

試題E ：七段碼

• dfs+并查集

// 80 #include <iostream> using namespace std;int res; int e[10][10]; bool vis[10]; int p[10];int find(int x) {if (p[x] != x) p[x] = find(p[x]);return p[x]; }bool check() {for (int i = 0; i < 10; i ++ ) p[i] = i;for (int i = 0; i < 10; i ++ )for (int j = 0; j < 10; j ++ )if (vis[i] && vis[j] && e[i][j]){int ii = find(i), jj = find(j);p[ii] = jj;}int cnt = 0;for (int i = 0; i <= 6; i ++ )if (vis[i] && p[i] == i)cnt ++ ;if (cnt != 1) return false;return true; }void dfs(int u) {if (u == 7){if (check()) res ++ ;return ;}vis[u] = true;dfs(u + 1);vis[u] = false;dfs(u + 1); }int main() {e[0][1] = e[1][0] = true;e[0][3] = e[3][0] = true;e[0][4] = e[4][0] = true;e[1][2] = e[2][1] = true;e[2][3] = e[3][2] = true;e[2][6] = e[6][2] = true;e[3][4] = e[4][3] = true;e[3][6] = e[6][3] = true;e[4][5] = e[5][4] = true;e[5][6] = e[6][5] = true;dfs(0);cout << res; }

試題F ：成績統計

#include <iostream> using namespace std;int main() {int n; cin >> n;int jige = 0, youxiu = 0;for (int i = 0; i < n; i ++ ){int grade; cin >> grade;if (grade >= 60) jige ++ ;if (grade >= 85) youxiu ++ ;}cout << (int)(100.0 * jige / n + 0.5) << '%' << endl;cout << (int)(100.0 * youxiu / n + 0.5) << '%' << endl; }

試題G ：回文日期

• 直接枚舉回文數，再來判斷合法性

#include <iostream> #include <algorithm> using namespace std;int months[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};int get_day(int yy, int mm) {if (mm != 2) return months[mm];if ((yy % 400 == 0) || (yy % 4 == 0 && yy % 100 != 0)) return 29;return 28; }int main() {int n; cin >> n;bool ok1 = false, ok2 = false;string ans1, ans2;for (int year = n / 10000; ; year ++ ){string a = to_string(year);string b = a;reverse(b.begin(), b.end());if (a + b <= to_string(n)) continue;int month = stoi(b.substr(0, 2));int day = stoi(b.substr(2, 2));if (month < 1 || month > 12) continue;if (day < 1 || day > get_day(year, month)) continue;if (!ok1) ans1 = a + b, ok1 = true;if (!ok2){if (year / 100 == year % 100)ans2 = a + b, ok2 = true;}if (ok1 && ok2) break;}cout << ans1 << endl << ans2; }

試題H ：子串分值和

• 算每一個位置的貢獻
• 因為是算每個連續子串中不同字符的個數，因此 對于原字符串中所有相同的字符而言，每一個字符做出的貢獻中，當前子串必然不包含上一次這個字符出現的位置及以前，包含這個字符出現的位置至原字符串末尾

#include <iostream> using namespace std;typedef long long ll;int last[1000];int main() {string s; cin >> s;int n = (int)s.size();s = "0" + s;ll res = 0;for (int i = 1; i <= n; i ++ ){res += (ll)(i - last[s[i] - 'a']) * (n - i + 1);last[s[i] - 'a'] = i;}cout << res; }

試題I ：平面切分

• 每增加一條直線，它對答案的貢獻是 它與之前直線的所有交點（要去重）個數+1
• 直線本身也需要去重
• 因此，兩次用到set
• 計算交點的時候，注意分母為0情況
• 易錯點 ：涉及直線或點，經常要用到set容器來去重，且每次計算分式必然要討論分母為零的情況

#include <iostream> #include <set> #include <vector> using namespace std;typedef pair<int, int> PII;set<PII> se1; vector<PII> linear;int main() {int n; cin >> n;for (int i = 0, a, b; i < n && scanf("%d%d", &a, &b); i ++ )se1.insert({a, b});for (auto pii : se1)linear.push_back(pii);int res = 2;for (int i = 1; i < linear.size(); i ++ ){set<pair<long double, long double>> point;for (int j = 0; j < i; j ++ ){int a1 = linear[i].first, b1 = linear[i].second;int a2 = linear[j].first, b2 = linear[j].second;if (a1 == a2) continue;long double x = (long double)(b2 - b1) / (a1 - a2);long double y = (long double)(a1 * b2 - a2 * b1) / (a1 - a2);point.insert({x, y});}res += point.size() + 1;}printf("%d", res); }

試題J ：字串排序

總結

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