蓝桥杯知识点对应的复习题

【注】

long long

字母大写变小写+=32

分离整数的各个位数

int main()
{int a; cin >> a;while (a>0){cout << a % 10 << " ";a /= 10;}return 0;
}

素数

试除法

bool check(int n)
{if (n < 2)return false;for (int i = 2; i <= sqrt(n); i++){if (n % i == 0)return false;}return true;
}
int main()
{int a, b; cin >> a >> b;for (int i = a; i <= b; i++){if (check(i))cout << i << endl;}return 0;
}

埃氏筛

原理：每次将素数的倍数全部标记上

const int N = 1e3 + 10;
bool vis[N];//标记数组
int prime[N];//素数表 0是素数 1不是素数
int id = 1;
//埃氏筛--O(nlonglongn)
void E_sieve(int n)
{//标记 0 1vis[0] = vis[1] = 1;for (int i = 2;i<=n; i++){if (vis[i] == 0)//是素数{prime[id++] = i;//筛掉素数i的所有倍数for (int j = i * i; j <= n; j += i){vis[j] = 1;}}}
}int main()
{int n; cin >> n;E_sieve(n);for (int i = 1; i < id; i++)cout << prime[i] << endl;return 0;
}

完全平方数

int main()
{int a; cin >> a;if (sqrt(a) == int(sqrt(a)))cout << "Y";elsecout << "N";return 0;
}

分解质因数

int main()
{int n; cin >> n;int i = 2, flag = 0;cout << n << "=";while (n != 1){if (n % i == 0){if (flag == 0){cout << i, flag = 1;}elsecout << "*" << i;n /= i;}elsei++;}return 0;
}

质因数的分解

int main()
{int n; cin >> n;for (int i = 2;; i++){if (n % i == 0){cout << n / i;break;}}return 0;
}

求第n小的质数

bool check(int n)
{if (n < 2) return false;for (int i = 2; i <= sqrt(n); i++){if (n % i == 0)return false;}return true;
}
int main()
{int n; cin >> n; int cnt = 0;for (int i = 2;; i++){if (check(i)){cnt++;if (cnt == n){cout << i;break;}}}return 0;
}

数组求最大值(第一个元素作为最大小值)

const int N = 1e3 + 10;
int a[N];
int main()
{int n; cin >> n;cin >> a[1]; int mx = a[1];int mxid = 1;for (int i = 2; i <= n; i++){cin >> a[i];if (a[i] > mx){mx = a[i];mxid = i;}}cout << mxid << endl;return 0;
}

***最长连续相同子序列-双指针版

#include<iostream>
#include<vector>
using namespace std;int main() {int n;  cin >> n;vector<int> v;for (int i = 0; i < n; i++) {int x; cin >> x;v.push_back(x);}v.push_back(0);int maxlen = 1, l = 0;for (int r = 1; r < v.size(); r++) {if (r == v.size()-1) {if (r - l > maxlen) maxlen = r - l;break;}if(v[l]!=v[r]){if (r - l > maxlen) maxlen = r - l;l = r;}}cout << maxlen << endl;return 0;
}

整数去重

const int N = 2e4 + 10,M=5e3+10;
int a[N],cnt[M];
int main()
{int n; cin >> n;for (int i = 1; i <= n; i++){cin >> a[i];if (cnt[a[i]] == 0)cout << a[i]<<" ";cnt[a[i]]++;}return 0;
}

矩阵乘法（三层for循环）

const int N = 1e2+10;
int a[N][N], b[N][N],c[N][N];
int main()
{int n, m, k; cin >> n>>m>>k;for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)cin >> a[i][j];for (int i = 1; i <= m; i++)for (int j = 1; j <= k; j++)cin >> b[i][j];//n*kfor (int i = 1; i <= n; i++){for (int j = 1; j <= k; j++){for (int k = 1; k <= m; k++){c[i][j] += a[i][k] * b[k][j];}cout << c[i][j]<<" ";}cout << endl;}return 0;
}

字符串排序

#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int N = 1e2 + 10;
//字符串数组
string a[N];
int main()
{int n; cin >> n;for (int i = 1; i <= n; i++){cin >> a[i];}sort(a+1,a+1+n);// 参1--数组首元素地址for (int i = 1; i <= n; i++)cout << a[i] << endl;return 0;
}

线性序列变环

int main()
{string s1; string s2;cin >> s1 >> s2;if (s1.length() < s2.length())swap(s1, s2);s1 += s1;if (s1.find(s2) != -1)cout << "true" << endl;elsecout << "false" << endl;return 0;
}

判断回文

反转函数

int main()
{string s,r;getline(cin, s);r = s;reverse(s.begin(), s.end());if (s == r)cout << "yes";elsecout << "no";return 0;
}

冒泡排序

const int N = 1e2 + 10;
int a[N];
/*冒泡排序*/
int main()
{int n; cin >> n;for (int i = 1; i <= n; i++)cin >> a[i];for (int j = n; j > 1; j--) {for (int i = 1; i < j; i++){if (a[i] < a[i + 1])swap(a[i], a[i + 1]);}}for (int i = 1; i <= n; i++){cout << a[i]<<endl;}return 0;
}

选择排序（最小交换次数）

const int N = 1e2 + 10;
int a[N];
int main() {int n;   cin >> n;for (int i = 1; i <= n; i++) cin >> a[i];for (int i = 1; i <= n; i++) {//默认第i个元素是最小的int maxId = i;for (int j = i + 1; j <= n; j++) {if (a[j] > a[maxId]) {maxId = j;}}swap(a[i], a[maxId]);}for (int i = 1; i <= n; i++) cout << a[i] << endl;return 0;
}

计数排序

const int N = 1e2 + 10;
int a[N],cnt[1010];
int main()
{int n; cin >> n;for (int i = 1; i <= n; i++){cin >> a[i];cnt[a[i]]++;}for (int i = 1000; i >= 0; i--){while(cnt[i]){cout << i << endl;cnt[i]--;}}return 0;
}

表达式问题--前中后

前缀表达式

double fun()
{string s; cin >> s;if (s[0] == '+')return fun() + fun();else if (s[0] == '-')return fun() - fun();else if (s[0] == '*')return fun() * fun();else if (s[0] == '/')return fun() / fun();else return stod(s);
}int main()
{printf("%lf", fun());return 0;
}

搜索与回溯

abc全排列

#include<iostream>
using namespace std;
const int N = 1e2 + 10;
string s;
bool vis[N];
char cnt[N];
void dfs(int dep)
{if (dep == s.size()){for (int i = 0; i < dep; i++)cout << cnt[i];cout << endl;return ;}for (int i = 0; i < s.size(); i++){if (!vis[i]){vis[i] = 1;cnt[dep] = s[i];dfs(dep + 1);vis[i] = 0;}}
}
int main()
{cin >> s;dfs(0);return 0;
}

组合问题

 #include<iostream>
using namespace std;
const int N = 10;
bool vis[N];
int cnt[N];
int n,r;
void dfs(int dep) {//5.终止条件+结果处理if (dep == r+1) {for (int i = 1; i < dep; i++) {printf("%3d", cnt[i]);}cout << endl;return;}//1.枚举方案for (int i = cnt[dep-1]+1; i <= n; i++) {//2.标记-防止重复搜索if (!vis[i]) {//3.搜索vis[i] = 1;cnt[dep] = i;dfs(dep+1);//4.回溯vis[i] = 0;}}
}
int main() {cin >> n >> r;;dfs(1);return 0;
}

广搜求迷宫问题最短路

char g[N][N];//迷宫数组
bool vis[N][N];//标记数组
int sx, sy, tx, ty;
int dx[] = { 0,0,1,-1 }, dy[] = { 1,-1,0,0 };
int n, m;
struct point { int x, y, step; };
int ans;
void bfs(point s) {queue<point> q;q.push(s);  vis[s.x][s.y] = 1;while (!q.empty()) {point cur = q.front();  q.pop();if (cur.x == tx && cur.y == ty) {ans= cur.step - 1 ;return;}for (int i = 0; i < 4; i++) {int bx = cur.x + dx[i], by = cur.y + dy[i];if (bx<1 || bx>n || by<1 || by>m) continue;if (g[bx][by] == '#') continue;if (vis[bx][by]) continue;q.push({ bx,by,cur.step + 1 });  vis[bx][by] = 1;}}
}
int main() {while (cin >> n >> m&&n&&m) {for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {cin >> g[i][j];if (g[i][j] == '@') sx = i, sy = j;if (g[i][j] == '*') tx = i, ty = j;}}ans = -1;memset(vis, 0, sizeof vis);bfs({ sx,sy,1 });cout << ans << endl;}return 0;
}