AtCoder A. AABCDDEFE
原题链接
中等
作者:
Coinisi.
,
2023-01-15 09:02:44
,
所有人可见
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阅读 173
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#define IOS std::ios::sync_with_stdio(false)
#define inf 0x3f3f3f3f
#define YES cout << "YES" << endl
#define NO cout << "NO" << endl
#define int long long
#define x first
#define y second
#define cmp [&](PII a, PII b){return a.y < b.y;}
const int N = 5e5+10, mod = 1e9+7, M = 5e7+5, K = 2e5+10, Z = 1e5+7, X = 3e5+3;
using namespace std;
typedef long long LL;
typedef priority_queue<int> PQI;
typedef priority_queue <int, vector<int>, greater<>> PQGI;
typedef pair<int, int> PII;
int a = 110000000;
void check1()
{
int t1 = a % 10;//3
int t2 = ((a % 1000) - (a % 100)) / 100;//3
if(t1 != t2)
{
if(t2 == 0) a = a - t1;
else a ++;
}
}
void check2()
{
// 998244353
int t1 = a / 1000; // 998244
int t2 = a / 10000; // 99824
int t3 = a / 100000;// 9982
t1 = t1 - t2 * 10;
t2 = t2 - t3 * 10;
if(t1 != t2) a = a + 10000;
}
void check3()
{
int t1 = a / 100000000;
int t2 = a / 10000000 - t1 * 10;
if(t1 != t2) a = a + 100000000;
}
void solve()
{
int n; cin >> n;
for(int i = 1; i < n; i ++)
{
a = a + 10;
check1();
check2();
check3();
}
cout << a << endl;
}
signed main()
{
IOS; int T = 1;
cin.tie(nullptr);
cout.tie(nullptr);
// cin >> T;
while( T -- ) solve();
return 0;
}