蓝桥杯历届试题集合

回文数字

题目链接

　　思路：DFS尝试每一位

#include <bits/stdc++.h>
using namespace std;
const int maxn =  2e5 + 10;
#define ll long long
#define ios std::ios::sync_with_stdio(false)
#define int long long
typedef unsigned long long ULL;
vector<int>ans;
int a[10];
void dfs(int step , int s)
{
    if(step == 6 || step == 7)
    {
        if(s == 0){
            bool ok = true;
            if(a[1] == 0)ok = false;
            for(int i = 1 ; i <= step / 2 ; i ++){
                if(a[i] != a[step - i]){
                    ok = false;
                    break;
                }
            }
            if(ok){
                int aa = 0;
                for(int i = 1 ; i <= step - 1 ; i ++){
                    aa = aa * 10 + a[i];
                }
                ans.push_back(aa);
            }
        }
        if(step == 7)return ;
    }
    for(int i = 0 ; i <= 9 ; i ++){
        if(i > s)break;
        a[step] = i;
        dfs(step + 1 , s - i);
    }
}



signed main()
{
    ios;
    cin.tie(0);///
    int n;
    cin >> n;
    dfs(1 , n);
    if(ans.size() == 0){
        cout << -1 << '\n';
    }
    else{
        sort(ans.begin() , ans.end());
        for(int i = 0 ; i < ans.size() ; i ++){
            cout << ans[i] << '\n';
        }
    }
    return 0;
}

 

翻硬币

题目链接

#include <bits/stdc++.h>
using namespace std;
const int maxn =  2e5 + 10;
#define ll long long
#define ios std::ios::sync_with_stdio(false)
#define int long long
typedef unsigned long long ULL;
#define debug(x) cout << #x << " : " << x << '\n';

vector<int>vec;
char a[maxn] , b[maxn];
signed main()
{
    ios;
    cin.tie(0);///
    cin >> a + 1 >> b + 1;
    int n = strlen(a + 1);
    for(int i = 1 ; i <= n ; i ++){
        if(a[i] != b[i])vec.push_back(i);
    }
    int ans = 0;
    for(int i = 0 ; i < vec.size() ; i += 2){
        ans += vec[i + 1] - vec[i];
    }
    cout << ans << '\n';
    return 0;
}

 

连号区间数

题目链接

思路：

　　看了一下觉得是个很难的题，没想到o(n²)可以过，大吃一斤...

　　在一个区间内如果这个区间的最大值减去最小值等于区间长度，即满足条件。

#include <bits/stdc++.h>
using namespace std;
const int maxn =  2e5 + 10;
#define ll long long
#define ios std::ios::sync_with_stdio(false)
#define int long long
typedef unsigned long long ULL;

int a[maxn];
signed main()
{
    ios;
    cin.tie(0);///
    int n;
    cin >> n;
    for(int i = 1 ; i <= n ; i ++){
        cin >> a[i];
    }
    int ans = n;
    for(int i = 1 ; i <= n ; i ++){
        int maxx = a[i] , minn = a[i];
        for(int j = i + 1 ; j  <= n ; j ++){
            maxx = max(a[j] , maxx) , minn = min(minn , a[j]);
            if(maxx - minn == j - i) ans ++;
        }
    }
    cout << ans << '\n';
    return 0;
}

 

 

 

买不到的数目

题目链接

思路： 

　　找规律找出来的

#include <bits/stdc++.h>
using namespace std;
const int maxn =  2e5 + 10;
#define ll long long
#define ios std::ios::sync_with_stdio(false)
#define int long long
typedef unsigned long long ULL;
signed main()
{
    ios;
    cin.tie(0);///
    int a , b;
    cin >> a >> b;
    cout << a * b - a - b << '\n';
    return 0;
}