Given Length and Sum of Digits
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
题意：给两个数m, s，求出位数等于m 且个位数字之和等于s的最小数和最大数。
解题思路：直接贪心的去构造。先判断输出-1，-1的情况：m*9
AC代码：
#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define INF 0x7fffffffint a[105], b[105];int main(){// freopen("in.txt","r",stdin); int m, s; while(scanf("%d%d", &m, &s)!=EOF) { if(!s && m==1) {printf("0 0\n"); continue; } //m == 1 && s == 0 if(s > m*9 || s == 0) { //构造不成功的情况 cout0; i--){ int x = min(ss, 9); b[i] = x; ss -= x; } if(ss) b[0] += ss; for(int i=0; i<m i x cout return> <br> <br> <p></p> <p><br> </p> <p><br> </p> <p><br> </p> </m></time.h></stdlib.h></math.h></string></map></set></queue></vector></algorithm></iostream></string.h></stdio.h>
