CF489C Given Length and Sum of Digits...

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;
    }
}

CF489C Given Length and Sum of Digits...

原文：https://www.cnblogs.com/Xiao-yan/p/14734624.html