目录
一、试题A
没啥好说的,这就是一个小学数学题,36 ×30 / 10 =108
答案:108
二、试题B
%1000就是要求 2^2023后三位,因为只是一个填空题,所有我们直接用计算器就可以求出来
答案:608
三、试题C
可以使用暴力或者写程序的方法进行求解
#include<bits/stdc++.h>
#define num first
#define y second
using namespace std;
typedef long long ll;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
// 十进制数转换为任意n进制数
// num:十进制数,n:转换成n进制数
string Itoa(int num,int n)
{
string str;
int rem;
char ch;
if(num == 0)
str = "0";
while(num > 0)
{
rem = num % n;
ch = (rem < 10) ? (rem + '0') : (rem - 10 + 'A');
str = ch + str;
num /= n;
}
return str;
}
// 计算每一位上的和
int sum(string str)
{
int sum = 0;
for(int i = 0;i < str.length();i ++)
{
sum += str[i] - '0';
}
return sum;
}
string str1;
string str2;
int main()
{
int cnt = 0;
int i = 1;
while(1)
{
str1 = Itoa(i,2);
str2 = Itoa(i,8);
// cout<<i<<' '<<str1<<' '<<str2<<endl;
if(sum(str1) == sum(str2))
{
cnt ++;
cout << "数字为" << i << endl;
cout << str1 << endl;
cout << str2 << endl;
cout << "cnt的值是" << cnt << endl;
}
if(cnt == 23)
{
cout << "找到了!!!:>" << i << endl;
cout << str1 << endl;
cout << str2 << endl;
break;
}
i ++;
}
return 0;
}
答案: 4169
四、试题D
约数个数模版题
#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
int primes[N], cnt;
bool st[N];
void init(int n)
{
for (int i = 2; i <= n; i ++)
{
if (!st[i]) primes[cnt ++] = i;
for (int j = 0; primes[j] * i <= n; j ++)
{
st[primes[j] * i] = true;
if (i % primes[j] == 0) break;
}
}
}
void divide(int x)
{
int res = 0;
for (int i = 0; i < cnt; i ++)
{
int p = primes[i];
if (x % p == 0)
{
while (x % p == 0)
{
res ++;
{
x /= p;
}
}
}
}
cout << res << endl;
}
int main()
{
init(N);
int m = 30;
while (m--)
{
int x;
cin >> x;
divide(x);
}
return 0;
}
答案: 901440
五、试题E
经典BFS(广度优先搜索)求联通块,由题意可知要求所有0的联通块中0的个数
#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef pair<int, int> PII;
const int N = 45;
char g[N][N];
bool st[N][N];
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, -1, 0, 1};
int ans;
void bfs(int x, int y)
{
queue<PII> q;
st[x][y] = true;
q.push({x, y});
ans++;
while (q.size())
{
auto t = q.front();
q.pop();
for (int i = 0; i < 4; i++)
{
int a = dx[i] + t.x, b = dy[i] + t.y;
if (a < 0 || a >= 30 || b < 0 || b >= 40) continue;
if (st[a][b]) continue;
if (g[a][b] != '0') continue;
ans++;
st[a][b] = true;
q.push({a, b});
}
}
}
int main()
{
for (int i = 0; i < 30; i++) cin >> g[i];
bfs(0, 0);
cout << ans << endl;
return 0;
}
答案:541
六、试题F
直接字符串,简单模拟即可,我只能说这道题就是一道经典送分题,如果这都做不出来的话那就是真白学了
#include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
cin >> s;
for(int i = 1; i < s.size(); i ++)
cout << s[i];
cout << s[1];
return 0;
}
七、试题G
这道题应该和上一道题的难度不相上下吧,就是一个简单的循环遍历字符串模拟,从后往前输出第一个元音字母即可
#include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
cin >> s;
for(int i = s.size() - 1; i >= 0; i--)
{
if(s[i] == 'a' || s[i] == 'o' || s[i] == 'e' || s[i] == 'i' || s[i] == 'u')
{
cout << s[i] << endl;
break;
}
}
return 0;
}
八、试题H
简单模拟即可
#include <bits/stdc++.h>
using namespace std;
int main()
{
long long int n;
cin >> n;
long long int sum = 11;
while(sum >= 10)
{
sum = 1;
while(n)
{
if((n % 10) != 0)
sum *= n % 10;
n /= 10;
}
cout << sum << " "<< endl;
n = sum;
}
return 0;
}
九、试题 I
BFS模版题,就是加了求最大公约数的步骤
#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef pair<int, int> PII;
const int N = 1010;
int g[N][N];
int n, m;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, -1, 0, 1};
int ans;
bool st[N][N];
int gcd(int a, int b)
{
return b ? gcd(b, a % b) : a;
}
void bfs(int x, int y)
{
queue<PII> q;
st[x][y] = true;
q.push({x, y});
ans++;
while (q.size())
{
auto t = q.front();
q.pop();
for (int i = 0; i < 4; i++)
{
int a = dx[i] + t.x, b = dy[i] + t.y;
if (a < 1 || a > n || b < 1 || b > m) continue;
if (st[a][b]) continue;
if (gcd(g[a][b], g[t.x][t.y]) <= 1) continue;
st[a][b] = true;
ans ++;
q.push({a, b});
}
}
}
int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
cin >> g[i][j];
int x, y;
cin >> x >> y;
bfs(x, y);
cout << ans << endl;
return 0;
}
十、试题 J
这个就是一个简单的滑动窗口,前缀和模版题,有一点需要注意,那就是开long long,要不然会超出范围
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n,k;
cin >> n >> k;
long long int s[100000];
for(int i = 1; i <= n; i ++)
{
cin >> s[i];
s[i] += s[i - 1];
}
long long int m;
for(int i = 1; i + k - 1 <= n; i ++)
{
m = max(s[i + k - 1] - s[i - 1],m);
}
cout << m << endl;
return 0;
}