1163 - Bank Robbery

1163 - Bank Robbery

 

In one very cold morning, Mark decides to rob a bank. But while trying hacking into the security system, he found that it is locked by some random value. He also found a pattern on the random number, that is if he chops off the last digit of a number A, he gets a new number B. Then he calculates (A-B). He checked the first few numbers of the security system which exactly equals (A-B). Being very excited to have found the pattern, he learns that there are like 500 levels on the security system. He calculated all those numbers by hand but took a lot of time. As a sign of his accomplishment he left a note on the vault stating the pattern. You were the first officer on the crime scene and you've obtained the note. So if you can figure out A from (A-B), you can rob the bank very quick!

By the way, Mark succeeded in robbing the bank but had a heart attack on the getaway car and crashed.

Input

Input starts with an integer T (≤ 500), denoting the number of test cases.

Each line contains a single positive integer between 10 and 10¹⁸ (inclusive), giving the value of A-B.

Output

For each case, print the case number and the possible values of A in ascending order. Separate consecutive numbers with a single space.

Sample Input	Output for Sample Input
4 31 18 12 17	Case 1: 34 Case 2: 19 20 Case 3: 13 Case 4: 18

题意：给出A-B，求A， B为A去掉最后一位形成的数。

分析：B = A / 10, 设A的最后一位为x， A-B的值为n（已知）， B * 10 + x - B = n, 即9 * B = n - x; 而 x的取值为0 到9. A = （n - x）* 10 / 9 + x。

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
typedef unsigned long long ll;///注意是无符号位长整型，有符号位长整型会溢出。
#define N 100
using namespace std;
ll n;

int main()
{
int T, cas;
ll a[N];

scanf("%d", &T);

cas = 0;

while(T--)
{
cas++;
memset(a, 0, sizeof(a));

scanf("%llu", &n);

int k = 0;
for(int i = 0; i <= 9; i++)
{
if((n - i) % 9 == 0)
a[k++] =(n-i)*10/9 + i;
}
sort(a, a+k);///输出要求按升序排列。

k = unique(a, a+k) - a;///类属性算法unique的作用是从输入序列中“删除”所有相邻的重复元素。
printf("Case %d:", cas);
for(int i = 0; i < k; i++)
printf(" %llu", a[i]);

printf("\n");

}
return 0;
}