蓝桥杯训练4.18
程序员文章站
2023-12-26 15:40:21
...
1.好好学习
#include<bits/stdc++.h>
using namespace std;
int main(){
int cnt = 0, ans = 0;
vector<int>w;
w.push_back(1);
w.push_back(1);
w.push_back(2);
w.push_back(3);
do{
if(w[0] == 1 && w[1] == 1 && w[2] == 2 && w[3] == 3)
ans++;
cnt++;
}while(next_permutation(w.begin(), w.end()));
int temp = __gcd(ans, cnt);
cout << ans / temp << "/" << cnt / temp << endl;
return 0;
}
答案:1/12
2.埃及分数
#include<bits/stdc++.h>
using namespace std;
int main(){
int ans = 0;
for(int i = 1; i < 5000; i++){
for(int j = i + 1; j < 5000; j++){
int t1 = i * j;
int t2 = i + j;
int temp = __gcd(t1, t2);
t1 /= temp;
t2 /= temp;
if(t2 == 2 && t1 == 45)
ans++;
}
}
cout << ans << endl;
return 0;
}
答案:7
注意a,b是不同的整数
3.金蝉素数
答案:k == 5
4.横向打印二叉树
比较恶心的模拟,先建树, 在打印图形。
#include<bits/stdc++.h>
using namespace std;
int n, le[200], ri[200], a[200];
void add(int x, int &node){
if(node == -1){
node = x;
le[x] = ri[x] = -1;
return ;
}
if(a[x] > a[node])
add(x, ri[node]);
else
add(x, le[node]);
return ;
}
char ans[1000][1000], row;
int fun1(char *ch, int x){
int len = 0;
while(x){
ch[len++] = x % 10 + '0';
x /= 10;
}
ch[len] = 0;
reverse(ch, ch + len);
return len;
}
int prin(int x, int &y, int node){
char num[10];
int len = fun1(num, a[node]), index_y = y, temp = y;
if(ri[node] != -1){
index_y = prin(x + len + 3, y, ri[node]);
for(int i = index_y + 1; i >= y; i--)
ans[i][x + len + 1] = '|';
ans[y][x + len + 2] = '-';
ans[index_y + 1][x + len] = '-';
y = index_y + 1;
temp = y;
}
for(int i = 0; i < len; i++){
ans[temp][x + i] = num[i];
}
if(le[node] != -1){
y++;
index_y = prin(x + len + 3, y, le[node]);
for(int i = temp; i <= y; i++)
ans[i][x + len + 1] = '|';
ans[y][x + len + 2] = '-';
ans[temp][x + len] = '-';
}
y = temp;
return max(temp, index_y);
}
void solve(){
for(int i = 0; i <= row; i++){
int len = 0;
for(int j = 0; j < 1000; j++){
if(ans[i][j])
len = j;
}
for(int j = 0; j <= len; j++){
if(ans[i][j])
printf("%c", ans[i][j]);
else
printf(".");
}
printf("\n");
}
return ;
}
int main(){
n = 1;
int head = -1;
while(cin >> a[n]){
add(n++, head);
}
// for(int i = 1; i <= n; i++)
// cout << le[i] << " " << ri[i] << endl;
int yy = 0;
row = prin(0, yy, 1);
solve();
return 0;
}
5.危险系数
O(n)暴力,并查集判断,总复杂度O(n * (n + m))
6.公式求值
比较难,参考这篇题解
https://www.cnblogs.com/gangduo-shangjinlieren/p/4372897.html