原题链接
- 题目:给了数字位数和数字位数之和,要构造出最大的数字和最小的数字。
- 题解:就是想着让 \(9\) 往后并且让 \(1\) 往前,然后中间用 \(0\),注意无效的时候是 \(s = 0\) 并且 \(m \neq 1\)
- 代码:
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N = 5e5 + 9;
const ll mod = 1e9 + 7;
bool vis[N];
char s[N];
int main() {
string ans1, ans2;
int m, s;
cin >> m >> s;
if ( s > 9 * m||s == 0) {
if (m == 1&&s == 0)cout << 0 << " " << 0 << endl;
else
cout <<"-1 -1\n";
}
else {
int S = s-1;
for (int i = 1; i <= m; i ++) {
int d = S;
if (d >= 9) d = 9;
S -= d;
ans1 = (char)(d + '0') + ans1;
}
ans1[0]++;
cout << ans1 << " ";
ans1[0]--;
reverse(ans1.bgin(), ans1.end());
for (int i = 0; i < ans1.size(); i ++) {
if (ans1[i] != '9') {
ans1[i] ++;break;
}
}
cout << ans1 << endl;
}
}