难度 1400
题目 Codeforces:
You have a positive integer m and a non-negative integer s. Your task is to find the smallest and the largest of the numbers that have length m and sum of digits s. The required numbers should be non-negative integers written in the decimal base without leading zeroes.
The single line of the input contains a pair of integers m, s (1?≤?m?≤?100,?0?≤?s?≤?900) — the length and the sum of the digits of the required numbers.
In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).
题目解析
没啥好说的，直接看代码就行了，题意很简单，就是你要给出两个数，分别是满足位数为m，各个位上数字加起来的和为s的所有数字中的最大值和最小值，如果不存在就输出“-1 -1”
1 #include<iostream> 2 using namespace std; 3 typedef long long ll; 4 int a[101], b[101]; 5 void makea(int m, int temp) 6 { 7 for (int i = m - 1; i >= 1; i--) 8 { 9 if (temp > 9) 10 { 11 a[i] = 9; 12 temp -= 9; 13 } 14 else if (temp > 1) 15 { 16 a[i] = temp - 1; 17 temp = 1; 18 } 19 else if (temp)a[i] = 0; 20 } 21 a[0] = temp; 22 } 23 void makeb(int m, int temp) 24 { 25 for (int i = 0; i < m; i++) 26 { 27 if (temp > 9) 28 { 29 b[i] = 9; 30 temp -= 9; 31 } 32 else if (temp > 0) 33 { 34 b[i] = temp; 35 temp = 0; 36 } 37 else if (!temp)b[i] = 0; 38 } 39 } 40 int main() 41 { 42 int m, s, temp; cin >> m >> s;//m是位数 s是和 43 if (m == 1 && s == 0)cout << "0 0"; 44 else if (m * 9 < s || s == 0)cout << "-1 -1"; 45 else 46 { 47 makea(m, s); 48 makeb(m, s); 49 for (int i = 0; i < m; i++)cout << a[i]; 50 cout << " "; 51 for (int i = 0; i < m; i++)cout << b[i]; 52 } 53 return 0; 54 }