AtCoder Beginner Contest 278 B - Misjudge the Time-优快云博客

本文链接：https://blog.youkuaiyun.com/aaa7888210/article/details/127945721

本文介绍了一个使用暴力模拟算法解决特定时间判断问题的例子。通过不断迭代检查直到找到满足条件的时间，展示了算法实现的具体步骤和逻辑。代码中包含了初始化参数、定义辅助函数及主循环等关键部分。

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暴力模拟即可。

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i = a;i<n;i++)
#define per(i,a,n) for(int i = n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
#define endl '\n'
typedef vector<int> VI;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) {return mrand()%x;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(int a,int b){return a*b/gcd(a,b);}

const int N=100010;
int n,m;
int a[N];

bool check(int x){
	int a=x/60,b=x%60;
	string s;
	if(a<10) s+="0";
	s+=to_string(a);
	if(b<10) s+="0";
	s+=to_string(b);
	swap(s[1],s[2]);
	a=stoi(s.substr(0,2)),b=stoi(s.substr(2));
	return a>=0&&a<=23&&b>=0&&b<=59;
}
 
int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m;
	int x=(n*60+m)%1440;
	while(1){
		if(check(x)) break;
		x=(x+1)%1440;
	}
	cout<<x/60<<" "<<x%60;
	return 0;
}