蓝桥杯刷题计划-洛谷-持续更新

P8598 [蓝桥杯 2013 省 AB] 错误票据

题目

#include <bits/stdc++.h>
#define endl '\n'
#define int long long 
#define INF 0x3f3f3f3f3f
const int N = 1000010;
using namespace std;
int arr[N];
signed main()
{
	int N;
	cin>>N;
	int idx;
	while(cin>>arr[++idx]){}
	//排序
	sort(arr+1,arr+idx+1);
	int ans1,ans2;
	for(int i = 1; i <= idx; i++){
		if(arr[i+1] - arr[i] == 2)
			ans1 = arr[i]+1;
		if(arr[i+1] == arr[i])
			ans2 = arr[i];
	
	}
	cout<<ans1<<' '<<ans2;
	return 0;
}

P8752 [蓝桥杯 2021 省 B2] 特殊年份

题目

#include<bits/stdc++.h>
#define ll longlong 

using namespace std;



int a[100010];
int main(){
	
	int n = 5;
	string s;
	int ans = 0;
	for(int i = 0; i < n;i++ ){
		cin>>s;
		if(s[0] == s[2]&& s[3] - s[1] == 1)
			ans++;
	}
	cout<<ans;
	
	return 0;
}

P8753 [蓝桥杯 2021 省 AB2] 小平方

题目

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;

const int N = 100010;
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	
	int n;
	cin>>n;
	int cnt= 0 ;
	for(int i = 1; i<= n-1; i++){
		int x = i*i;
		if((x % n) < n/2.0){
			cnt++;
		}
	}
	cout<<cnt;
	return 0;
}

P8780 [蓝桥杯 2022 省 B] 刷题统计

题目

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;

const int N = 100010;
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	
	int a,b,n;
	cin>>a>>b>>n;
	int sum = 0;
	int cnt = 1;
	while(sum<n){
		if(cnt % 7>=1&& cnt%7<=5)
			sum+=a;
		else
			sum+=b;
		cnt++;
	
	}
	
	cout<<cnt-1;
	return 0;
}

P8706 [蓝桥杯 2020 省 AB1] 解码

题目

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;

const int N = 100010;
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	
	string s;
	cin>>s;
	for(int i = 0;i < s.size();i++){
		if(s[i]>='1'&& s[i]<='9')
			for(int j = 0 ; j < s[i]-'1';j++)
				cout<<s[i-1];
		else
			cout<<s[i];
	}
	return 0;
}

P8717 [蓝桥杯 2020 省 AB2] 成绩分析

题目

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;

const int N = 100010;
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	
	int n;
	int a[N];
	cin>>n;
	for(int i = 0 ;i < n;i++){
		cin>>a[i];
	}
	
	sort(a,a+n);
	double sum = 0;
	for(int i = 0 ; i < n;i++)
		sum+= a[i];
	double ans = sum/n*1.0;
	cout<<a[n-1]<<endl<<a[0]<<endl;
	printf("%.2lf",ans);
	return 0;
}

P8711 [蓝桥杯 2020 省 B1] 整除序列

题目

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;

const int N = 100010;
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n ;
	cin>>n;
	while(n>0){
		cout<<n<<" ";
		n = n>>1;
	}
	return 0;
}

P8722 [蓝桥杯 2020 省 AB3] 日期识别

题目

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;
	unordered_map<string,int> mp{//创立哈希表，建立字符到整形的映射
	{"Jan",1},
	{"Feb",2},
	{"Mar",3},
	{"Apr",4},
	{"May",5},
	{"Jun",6},
	{"Jul",7},
	{"Aug",8},
	{"Sep",9},
	{"Oct",10},
	{"Nov",11},
	{"Dec",12}
};
const int N = 100010;
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	
	string s;
	cin>>s;
	string s1 = s.substr(0,3);
	string s2 = s.substr(3);
	
	cout<<mp[s1]<<" ";
	if(s2[0] != '0')
		cout<<s2[0];
	
	cout<<s2[1];
	return 0;
}

P8680 [蓝桥杯 2019 省 B] 特别数的和

题目

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;


const int N = 100010;
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	
	int n;
	int ans = 0;
	cin>>n;
	for(int i = 1;i<=n;i++){
		string s = to_string(i);
		for(int j =0; j <s.size();j++){
			if(s[j] == '0' || s[j] =='1'||s[j] =='2'||s[j] == '9')
			{	ans+=i;
				break;
				}
		}
	}
	cout<<ans;
	return 0;
}

P9240 [蓝桥杯 2023 省 B] 冶炼金属

题目

思路贪心

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;


const int N = 100010;
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	
	int n;
	cin>>n;
	int a,b;
	int ans_min = 0,ans_max = 1e9;
	for(int i= 0; i< n;i++){
		cin>>a>>b;
		ans_min = max(ans_min,a/(b+1)+1);
		ans_max = min(ans_max,a/b);
	}
	cout<<ans_min<<" "<<ans_max;
	return 0;
}

P8597 [蓝桥杯 2013 省 B] 翻硬币

题目

思路模拟
遇到不同的就反转

#include <bits/stdc++.h>
#define int long long 
#define endl '\n'
#define INF 0x3f3f3f3f3f
using namespace std;
const int N = 100010;
int arr[N];

signed main(){

    string a,b;
    cin>>a>>b;
    int cnt = 0;
    for(int i = 0 ; i < a.size();i++){
        if(a[i]!=b[i]){
            a[i] = a[i] == 'o'?'*':'o';
            a[i+1] = a[i+1] =='o'?'*':'o';
            cnt++;
        }

    }
    cout<<cnt;

    return 0;
}

P8647 [蓝桥杯 2017 省 AB] 分巧克力

思路打暴力超时题干字眼——输出切出的正方形巧克力最大可能的边长 二分答案
对于第 i 块巧克力，当边长为 x 时，可以分出 ⌊(ai÷x)×(bi÷x)⌋ 块巧克力。

题目

#include <bits/stdc++.h>
#define int long long 
#define endl '\n'
#define INF 0x3f3f3f3f3f
using namespace std;
const int N = 100010;
int a[N],b[N];//长宽
int n,k;

bool check(int mid){
	int cnt =0 ;
	for(int i = 1 ; i <= n ;i++){
		cnt+=(a[i]/mid)*(b[i]/mid);
	}
	if(cnt>=k)return true;
	else return false;
}

signed main(){
	
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n>>k;
	for(int i = 1; i <= n;i++){
		cin>>a[i]>>b[i];
	}
	int l = 0;
	int r = 10010;
	while(l+1<r){
		int mid = (l+r)>>1;
		if(check(mid))
			l = mid;
		else
			r = mid;
	}
	if(check(r)) cout<<r;
	else cout<<l;
	return 0;
}

P8772 [蓝桥杯 2022 省 A] 求和

题目

思路 前缀和注意数据范围！
S=a1⋅a2+a1⋅a3+⋯+a1⋅an+a2⋅a3+⋯+an−2⋅an−1+an−2⋅an+an−1⋅an

=(a2+a3+⋯+an)⋅a1+(a3+a4+⋯+an)⋅a2+⋯+an⋅an−1

#include <bits/stdc++.h>
#define int long long 
#define endl '\n'
#define INF 0x3f3f3f3f3f
using namespace std;
const int N = 200010;
int arr[N],sum[N];
int n;
int ans;
signed main(){
	
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	for(int i = 1 ; i<=n;i++){
		cin>>arr[i];
		sum[i] = sum[i-1] + arr[i];
	}
	for(int i =1 ;i <=n-1 ;i++){
		ans += arr[i]*(sum[n]-sum[i]); 
	}
	cout<<ans;
	return 0;
}

P8665 [蓝桥杯 2018 省 A] 航班时间

题目

前置知识

- 去程时间 =飞行时间 + 时差；回程时间 = 飞行时间 − 时差
- 由此可知飞行时间 = （去+回）/ 2

用 scanf 输入前面的时间：

scanf("%d:%d:%d %d:%d:%d",&h1,&m1,&s1,&h2,&m2,&s2);

每一行后面不一定有额外的天数。如果有，则中间一定有空格，所以判断下一个字符是否为空格即可。
为方便计算，不妨把时间都转换成以秒为单位 t=86400⋅d+3600⋅h+60⋅m+s
h = ans/3600 min = ans%3600/60 s = ans%60

#include <bits/stdc++.h>
#define endl '\n'
#define INF 0x3f3f3f3f3f
const int N = 1000010;
using namespace std;

int get()
{
    int h1,m1,s1,h2,m2,s2,day=0;
    scanf("%d:%d:%d %d:%d:%d",&h1,&m1,&s1,&h2,&m2,&s2);
    if(getchar()==' ')scanf("(+%d)",&day);
    return (day*86400+h2*3600+m2*60+s2)-(h1*3600+m1*60+s1);
}
signed main()
{
    int T;
    scanf("%d",&T);
    while(T--)
        {
            int ans=(get()+get())/2;
            printf("%02d:%02d:%02d\n",ans/3600,ans%3600/60,ans%60);
        }
    return 0;
}

P8681 [蓝桥杯 2019 省 AB] 完全二叉树的权值

题目

思路模拟
完全二叉树的性质

设深度为 dep，根节点的深度为 1。则有第 dep 层的节点为 2dep，每层开头的节点编号为 2dep−1，末尾的节点编号为 2dep−1（以上结论叶子节点除外）。

注意上面加粗黑体字第一次没考虑到错了两个点 叶子节点需要额外特判

#include <bits/stdc++.h>
#define int long long 
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;
const int N = 100010;
int arr[N];
int n;
int dep = 1,sum = 0,Max = -1,a,ans;
signed main(){
	
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	for(int i = 1; i <= n; i++){
		cin>>a;
		sum+=a;
		if(i == (1<<dep)-1){//判断当前位置是否是该dep的最后一个节点？结算:go on 
			if(sum>Max){
				Max = sum;
				ans = dep;
			}
		dep++;	
		sum = 0;
		}
	}
	if(sum>Max){//叶子节点特判
		Max = sum;
		ans = dep;
	}
	
	cout<<ans;
	return 0;
}

P9231 [蓝桥杯 2023 省 A] 平方差

题目

思路 找规律
参考题解

#include <bits/stdc++.h>
#define int long long 
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;
const int N = 100010;
int arr[N];
int n;
int f(int x) {//小于等于x的奇数个数 
	if (!x) return 0;
	return (x + 1) / 2;
}
int g(int x) {//小于等于x的4的倍数个数 
	return x / 4;
}
signed main(){
	
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int l, r; cin >> l >> r;
	cout << f(r) - f(l - 1) + g(r) - g(l - 1);
	return 0;
}

P9230 [蓝桥杯 2023 省 A] 填空问题

题目

思路
problem A 填空题 本地暴力枚举
把数字转成字符串去处理
答案为 4430091

#include <bits/stdc++.h>
#define int long long 
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;
const int N = 1e8;
bool check(string s){
	int ans1 = 0 ,ans2 = 0;
	for(int i = 0; i < s.size();i++){
		if(i<s.size()/2)
			ans1 += s[i]-'0';
		else
			ans2 += s[i]-'0';
	}
	return ans1 == ans2;
}
signed main(){
	
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	string s;
	int ans = 0;
	for(int i = 1 ; i<= N;i++){
		s= to_string(i);
		if(s.size()%2 == 1)continue;//奇数
		if(check(s)) {
			ans++;
		}
		
	}
	cout<<ans;
	return 0;
}

problem B DFS 本地暴搜
答案为 4165637

#include <bits/stdc++.h>
#define int long long 
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;
const int N = 1e8;
int arr[N];
int ans;
void dfs(int score,int cnt){
	if(cnt>30||score == 100)
		return;
	if(cnt == 30 && score == 70)
		ans++;
	dfs(score+10,cnt+1);
	dfs(0,cnt+1);
}


signed main(){
	
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	//表示0分 0题
	dfs(0,0); 
	cout<<ans;
	return 0;
}

P8649 [蓝桥杯 2017 省 B] k 倍区间

题目

题解参考

思路看到”连续子序列求和”这一要求时，我们果断选择前缀和解答

#include <bits/stdc++.h>
#define int long long 
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;
const int N = 100010;
int arr[N],sum[N];
int n,k;
map<int,int> mp;

signed main(){
	
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n>>k;
	int ans = 0;
	mp[0] = 1;
	for(int i = 1; i<=n;i++){
		cin>>arr[i];
		sum[i] = (sum[i-1]+ arr[i])%k;//记录前缀和模k的值
		mp[sum[i]]++;
	}
	for(int i = 0; i< n;i++){
		ans+=(mp[i]*(mp[i]-1))/2;
	}
	cout<<ans;
	return 0;
}

自己的理解
解释：1要和其他1组一起满足

0可以单独算也可以组在一起满足

#include <bits/stdc++.h>
#define int long long 
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;
const int N = 100010;
int arr[N],sum[N];
int n,k;
map<int,int> mp;

signed main(){
	
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n>>k;
	int ans = 0;
	for(int i = 1; i<=n;i++){
		cin>>arr[i];
		sum[i] = (sum[i-1]+ arr[i])%k;//记录前缀和模k的值
		mp[sum[i]]++;
	}
	for(auto i :mp){
		if(i.first == 0)
			ans+=i.second*(i.second+1)/2;
		else
			ans+=i.second*(i.second-1)/2;
	}
	cout<<ans;
	return 0;
}