Old Bill
Time Limit: 2 Seconds Memory Limit: 65536 KB
Among grandfather��s papers a bill was found:
72 turkeys $_679_
The first and the last digits of the number that obviously represented the total price of those turkeys are replaced here by blanks (denoted _), for they are faded and are now illegible. What are the two faded digits and what was the price of one turkey?
We want to write a program that solves a general version of the above problem:
N turkeys $_XYZ_
The total number of turkeys, N, is between 1 and 99, including both. The total price originally consisted of five digits, but we can see only the three digits in the middle. We assume that the first digit is nonzero, that the price of one turkey is an integer number of dollars, and that all the turkeys cost the same price.
Given N, X, Y , and Z, write a program that guesses the two faded digits and the original price. In case that there is more than one candidate for the original price, the output should be the most expensive one. That is, the program is to report the two faded digits and the maximum price per turkey for the turkeys.
Input
The input consists of T test cases. The number of test cases (T) is given on the first line of the input file. The first line of each test case contains an integer N (0 < N < 100), which represents the number of turkeys. In the following line, there are the three decimal digits X, Y , and Z, separated by a space, of the original price $_XYZ_.
Output
For each test case, your program has to do the following. For a test case, there may be more than one candidate for the original price or there is none. In the latter case your program is to report 0. Otherwise, if there is more than one candidate for the original price, the program is to report the two faded digits and the maximum price per turkey for the turkeys. The following shows sample input and output for three test cases.
Sample Input
3
72
6 7 9
5
2 3 7
78
0 0 5
Sample Output
3 2 511
9 5 18475
0
很简单的一道题目，开始的时候是我想多了，原来最简单的两重循环就可以满足效率要求了。原先我还想着是不是要根据N的值先确定总价格的尾数，然后再进行循环呢，原来不用这么麻烦啊。
#include<iostream>
using namespace std;
int main()
{
int cases;cin>>cases;
while(cases--)
{
int n,x,y,z;
cin>>n>>x>>y>>z;
bool breakFor = false;
for(int thousand = 9; thousand >= 1; thousand--)
{
breakFor = false;
for(int unit = 9; unit >= 0; unit--)
{
int destination = thousand*10000 + x*1000 + y*100 + z*10 + unit;
if(destination % n == 0)
{
cout<<thousand<<""<<unit<<""<<destination/n<<endl;
breakFor = true;
break;
}
}
if(breakFor)
break;
}
if(!breakFor)
cout<<0<<endl;
}
}