2-1基础算法-枚举/模拟

文章目录

1.枚举
2.模拟

1.枚举

[例1] 特别数的和
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评测系统

#include <iostream>
using namespace std;
bool pa(int x) {while (x) {if (x % 10 == 2 || x % 10 == 1 || x % 10 == 0 || x % 10 == 9) {return true;}else {x = x / 10;}}return false;
}
int main()
{int sum=0;int m;cin >> m;for (int i = 1; i <= m; i++) {if (pa(i)) {sum += i;}}cout << sum;return 0;
}

[例2] 反序数
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评测系统

#include <iostream>
using namespace std;
int main()
{int n;cin >> n;int a, b, c;cin >> a >> b >> c;int sum = 0;for (int i = 1; i <= n; i++) {if (i % a != 0 && i % b != 0 && i % c != 0) {sum++;}}cout << sum;return 0;
}

[例3] 找到最多的数
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评测系统

#include <iostream>
#include <vector>
#include <map>
using namespace std;
int main()
{int n, m;cin >> n >> m;int num = n * m;map<int,int> mp;vector<int> a(num);//记录去重后的序列while (num--) {int x;cin >> x;if (!mp.count(x)) {a.push_back(x);}mp[x]++;}for (const auto& x:a) { //也可以不使用vector，每次从map中取出[x.y]if (mp[x] * 2 > n * m)cout << x;}
}

[例4]
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评测系统

对于每种可能的颜色，我们尝试计算如果将走廊涂成这种颜色需要多少天。在遍历过程中要跳过已经是这种颜色的房子，每次移动k步。最后记录对于这种颜色所需的涂色天数，并与之前的最小天数进行比较，取较小值。
输入数据的最大值即为初始颜色种类数

#include <iostream>
#include <vector>
using namespace std;
int main()
{int t;cin >> t;while (t--) {int n, k;cin >> n >> k;vector<int> a(n);int maxnum = 0;//记录初始颜色种类数for (int i = 0; i < n; ++i) {cin >> a[i];maxnum=max(maxnum, a[i]);}int ans = n;for (int c = 1; c <= maxnum; ++c) {int cnt = 0;int idx = 0;while (idx < n) {if (a[idx] == c) {idx++;continue;}idx += k;cnt++;}ans = min(cnt, ans);}cout << ans << endl;}
}

2.模拟

[例1] 扫雷
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评测系统

#include <iostream>
using namespace std;
const int N = 150;
int main()
{int n, m;cin >> n >> m;int a[N][N], ans[N][N];for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {cin >> a[i][j];}}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (a[i][j] == 1) { ans[i][j] = 9; continue; }int i2 = i, j2 = j;//遍历四周for (i2 = max(0, i - 1); i2 <= min(n - 1, i + 1); i2++) {for (j2 = max(0, j - 1); j2 <= min(m - 1, j + 1); j2++) {if (a[i2][j2] == 1) ans[i][j]++;}}}}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {cout << ans[i][j]<<" ";}cout << endl;}return 0;
}

[例2] 灌溉
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评测系统

使用类似于BFS的过程，用bool类型的irrigated数组记录当前位置是否已被灌溉，最终统计已被灌溉的数量。

#include <iostream>
#include <queue>
using namespace std;
const int N = 200;
int main()
{int n, m, t;cin >> n >> m >> t;bool irrigated[N][N];queue<pair<int, int>> q;for (int i = 0; i < t; i++) {int r, c;cin >> r >> c;r--;c--;irrigated[r][c] = true;q.push({ r,c });}int k;cin >> k;for (int minute = 0; minute < k; ++minute) {int size = q.size();for (int i = 0; i < size; ++i) {pair<int,int > fr=q.front();//取到水管的位置坐标int x = fr.first;int y = fr.second;q.pop();for (int i2 = max(0,x - 1); i2 <= min(n,x + 1); i2++) { //四周灌溉for (int j2 = max(0,y - 1); j2 <= min(m,y + 1); j2++) {if (!irrigated[i2][j2]) {irrigated[i2][j2] = true;q.push({i2, j2});}}}}}int count = 0;for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {if (irrigated[i][j]) {count++;}}}cout << count;return 0;
}

也可使用map

#include <iostream>
#include <map>
using namespace std;
const int N = 200;
int main()
{int n, m, t;cin >> n >> m >> t;bool irrigated[N][N];map<int, int> q;for (int i = 0; i < t; i++) {int r, c;cin >> r >> c;r--;c--;irrigated[r][c] = true;q[r]=c;}int k;cin >> k;for (int minute = 0; minute < k; ++minute) {int size = q.size();for (const auto& i:q) {int x = i.first;int y = i.second;for (int i2 = max(0,x - 1); i2 <= min(n,x + 1); i2++) {for (int j2 = max(0,y - 1); j2 <= min(m,y + 1); j2++) {if (!irrigated[i2][j2]) {irrigated[i2][j2] = true;q[i2]=j2;}}}}}int count = 0;for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {if (irrigated[i][j]) {count++;}}}cout << count;return 0;
}

[例3] 回文日期

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评测系统

#include <iostream>
#include <string>
using namespace std;
bool huiwen(string s) {if (s[0] == s[7] && s[1] == s[6] && s[2] == s[5] && s[3] == s[4]) {return true;}elsereturn false;
}
bool er(string s) {if (s[0] == s[2] && s[2] == s[5] && s[5] == s[7] && s[1] == s[3] && s[3] == s[4] && s[4] == s[6]) {return true;}elsereturn false;
}
int pdday(int year, int month) {if (month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12)return 31;else if (month == 2) {if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0) {return 29;}elsereturn 28;}elsereturn 30;
}
string zifuchuan(int year, int month, int day) {string s=to_string(year);if (month / 10 == 0) {s += '0' + to_string(month);}else {s += to_string(month);}if (day / 10 == 0) {s += '0' + to_string(day);}else {s += to_string(day);}return s;
}
int main()
{int day, month, year;int chushi;int day1,day2;cin >> chushi;day1 = chushi % 10;chushi = chushi / 10;day2= chushi % 10;chushi = chushi / 10;day=day2 * 10 + day1;day1 = chushi % 10;chushi = chushi / 10;day2 = chushi % 10;chushi = chushi / 10;month = day2 * 10 + day1;year = chushi;bool flag = 0;bool pdhuiwen = 0;bool pder = 0;int i, j, k;for (i = year;; i++) {for (j = 1; j <= 12; j++) {if (flag == 0) {j = month;}for (k = 1; k <= pdday(year,month); k++) {if (flag == 0) {k = day+1;flag = 1;}string s=zifuchuan(i, j, k);if (pdhuiwen!=1&&huiwen(s)) {cout << s << endl;pdhuiwen = 1;}if (pder!=1&&er(s)) {cout << s;pder = 1;}if (pdhuiwen == 1 && pder == 1) {return 0;}}}}
}

[例4] 小蓝和小桥的挑战
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评测系统

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{int n;cin >> n;while (n--) {int m;cin >> m;vector<int> a(m);for(int i=0;i<m;i++){cin >> a[i];}int cnt = count(a.begin(), a.end(), 0); // #include <algorithm>int sum = 0;for (const auto& x : a) {sum += x;}if (sum == 0) {cout << cnt + 1;}elsecout << cnt<<endl;}
}

[例5] DNA序列修正

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评测系统

#include <iostream>
using namespace std;
bool pipei(char a, char b) {return (a == 'A' && b == 'T') || (a == 'T' && b == 'A') || (a == 'C' && b == 'G') || (a == 'G' && b == 'C');
}
int main()
{int n;cin >> n;string s1, s2;cin >> s1 >> s2;int cnt = 0;//计数for (int i = 0; i < n; ++i) {if (!pipei(s1[i], s2[i])) {cnt++;for (int j = i + 1; j < n; j++) {if (!pipei(s1[j], s2[j])&& pipei(s1[i], s2[j])&& pipei(s1[j], s2[i])) {swap(s2[i], s2[j]);break;}}}}cout << cnt;
}

[例6] 无尽的石头
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评测系统

#include <iostream>
using namespace std;
int ssum(int x) {int sum = 0;while (x) {sum += x % 10;x /= 10;}return sum;
}
int main()
{int t;cin >> t;while (t--) {int x = 0;cin >> x;int step = 0;bool end = 0;for (int i = 1; i <= x; i += ssum(i)) {if (i == x) {cout << step << endl;end = 1;}step++;}if (end == 0) {cout << -1 << endl;}}
}