USER: tianbing tianbing [tbbd4261]
TASK: crypt1
LANG: C++
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.011 secs, 2928 KB]
Test 2: TEST OK [0.011 secs, 2928 KB]
Test 3: TEST OK [0.022 secs, 2928 KB]
Test 4: TEST OK [0.000 secs, 2928 KB]
Test 5: TEST OK [0.000 secs, 2928 KB]
Test 6: TEST OK [0.011 secs, 2928 KB]
Test 7: TEST OK [0.011 secs, 2928 KB]
All tests OK.
YOUR PROGRAM ('crypt1') WORKED FIRST TIME! That's fantastic
-- and a rare thing. Please accept these special automated
congratulations.
Here are the test data inputs:
------- test 1 -------
5
2 3 4 6 8
------- test 2 -------
4
2 3 5 7
------- test 3 -------
1
1
------- test 4 -------
7
4 1 2 5 6 7 3
------- test 5 -------
8
9 1 7 3 5 4 6 8
------- test 6 -------
6
1 2 3 5 7 9
------- test 7 -------
9
1 2 3 4 5 6 7 8 9
Keep up the good work!
Thanks for your submission!
簡單就好，沒考慮效率；
題目限制：三位數乘以兩位數，三位數乘以兩位數的個位是個三位數（part1），同樣三位數乘以兩位數的十位也是一個三位數（part2）。
結果是個四位數，并且所有的數字都在題目給的范圍中；
笨的方法：
three數組存所有可能的三位數，two存所有可能的兩位數，判斷part1 part2 以及最后的結果result即可
1
/**//*
2
ID:tbbd4261
3PROG:crypt1
4LANG:C++
5
*/
6
7#include<fstream>
8#include<vector>
9using namespace std;
10int a[10]=
{0};
11
12bool valid(int x,int n) //判斷數字是否合法
13{
14 for(int i=1; i<=n; i++)
15 if(a[i]==x)return true;
16 return false;
17}
18
19bool ispri(int x,int n)//判斷數是否由給定數字組成
20{
21 while(x)
22
{
23 int i=x%10;
24 if(!valid(i,n))return false;
25 x=x/10;
26
}
27 return true;
28}
29
30int main()
31{
32 ifstream fin("crypt1.in");
33 ofstream fout("crypt1.out");
34 int n,k,i,j,cnt=0;
35 int result;
36 vector<int> three;
37 vector<int> two;
38 fin>>n;
39 for(i=1; i<=n; i++)
40 fin>>a[i];
41 for(i=1; i<=n; i++)
42 for(j=1; j<=n; j++)
43 {
44 two.push_back(a[i]*10+a[j]);
45 for(k=1; k<=n; k++)
46 three.push_back(a[i]*100+a[j]*10+a[k]);
47 }
48
49 for(i=0; i<three.size(); i++)
50 for(j=0; j<two.size(); j++)
51 {
52 int part1=three[i]*(two[j]%10);
53 int part2=three[i]*(two[j]/10);
54 result=three[i]*two[j];
55 if(result>=1000&&result<10000&&ispri(result,n)
56 &&part1>=100&&part1<=999&&part2>=100&&part2<=999&&ispri(part1,n)&&ispri(part2,n))
57 cnt++;
58 }
59 fout<<cnt<<endl;
60
61 return 0;
62}