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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).
2 15
69 96
3 0
-1 -1
/*
贪心
对于min第一位先保证有1从后面开始加，如果加完之后还有s，说明NO，如果连1都没有那么NO
对于max就从头往后如果s够就9不够就s
其他特殊情况要判，比如1 0 坑点应该输出0 0。。其他带有0的都是-1 -1
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
int n,s;
int a[110];
while(~scanf("%d%d", &n, &s)){
if(n == 1 && s == 0) {
printf("0 0\n");
continue;
}
if(n == 0 || s == 0 ){
if( n== 1 && s == 0);
else {
printf("-1 -1\n");
continue;
}
}
memset(a, 0, sizeof(a));
int temp = s;
s --;
a[1] = 1;
int pos = n;
int flag = 0;
while(s > 0){
if(s <= 8){
a[pos--] += s;
s = 0;
}
else{ a[pos--] = 9;
s -= 9;
}
if(pos == 0) break;
}
int flag1 = 0;
int sum1 = 0;
for(int i = 1; i <= n; i++){
sum1 += a[i];
if(a[i] > 9)
flag1 = 1;
}
if(sum1 != temp || flag1 == 1) printf("-1 ");
else {
for(int i = 1; i <= n ;i++)
printf("%d", a[i]);
printf(" ");
}
memset(a, 0, sizeof(a));
s = temp;
for(int i = 1; i <= n ;i++){
if(s <= 8) {a[i] = s;s = 0;}
else {a[i] = 9; s -= 9; }
}
int sum = 0;
for(int i = 1; i <= n ; i++)
sum += a[i];
if(sum != temp || a[1] == 0) printf("-1\n");
else {
for(int i = 1; i <= n ;i++)
printf("%d",a[i]);
puts("");
}
}
return 0;
}
Codeforces Round #277.5 (Div. 2)——C贪心—— Given Length and Sum of Digits
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原文地址：http://www.cnblogs.com/zero-begin/p/4652382.html
