Codeforces Round #277.5 (Div. 2)(C题)_html/css_WEB-ITnose
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
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).
Sample test(s)
input
2 15
output
69 96
input
3 0
output
-1 -1
#include#include #include using namespace std;bool can(int m, int s){ if(s >= 0 && 9*m >= s) return true; else return false;}int main(){ int m,s; cin>>m>>s; if(!can(m,s)) { cout= 10) { cout 0 || (j == 0 && i > 1) ) && can(m - i, sum - j)) { minn += char('0' + j); sum -= j; break; } } sum = s; for(int i = 1; i = 0; j--) { if(can(m - i, sum - j)) { maxn += char('0' + j); sum -= j; break; } } cout
上一篇: Java循环
下一篇: php面向对象编程之封装性
推荐阅读
-
Codeforces Round #649 (Div. 2) C-Ehab and Prefix MEXs
-
Codeforces Round #432 (Div. 2) - C - Five Dimensional Points
-
Educational Codeforces Round 60 (Rated for Div. 2) ----A - Best Subsegment(思维题)
-
构造思维+树形结构 Codeforces Round #612 (Div. 2) D题 Numbers on Tree
-
Codeforces Round #658 (Div. 2)C1. Prefix Flip (Easy Version)(贪心)
-
Codeforces Round #658 (Div. 2) (C1、C2)
-
Codeforces Round #656 (Div. 3) (C、D题)
-
Codeforces Round #320 (Div. 2) C. A Problem about Polyline ( 数学 )
-
Codeforces Round #659 (Div. 2) C、String Transformation 1(思维+set)
-
Codeforces Round #654 (Div. 2)-C. A Cookie for You