C. Given Length and Sum of Digits...
time limit per test
1 secondmemory limit per test
256 megabytesinput
standard inputoutput
standard outputYou 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.
Input
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.
Output
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).
Examples
input
2 15
output
69 96
input
3 0
output
-1 -1
这道题目的要求大致是输入位数m和位数和s,输出最小数和最大数,没有就输出两个-1。
基本思路是高位先用9,然后低位补齐,最小数倒过来最高位用1,低位用9。
#include<iostream> #include<cstdlib> using namespace std; int main() { int m, s, i, j; cin >> m >> s; if (m == 1 && s < 10) { cout << s << ' ' << s << endl; return 0; } else if (s == 0||(m * 9 < s)) { cout << -1 << ' ' << -1 << endl; return 0; } int a[105] = {0}; int sum = s; for (i = 0;i < m;i++) { if (sum <= 9) { a[i] = sum; j = i; break; } a[i] = 9; sum -= 9; } if (a[m - 1] != 0) for (i = m - 1;i >= 0;i--) cout << a[i]; else { cout << 1; for (i = m - 2;i >= 0;i--) { if (i == j) { cout << a[i] - 1; continue; } cout << a[i]; } } cout << ' '; for (i = 0;i < m;i++) cout << a[i]; cout << endl; return 0; }