problem in Light OJ:
1001 - Opposite Task:
This problem gives you a flavor the concept of special judge. That means the judge is smart enough to verify your code even though it may print different results. In this problem you are asked to find the opposite task of the previous problem.
To be specific, I have two computers where I stored my problems. Now I know the total number of problems is n. And there are no duplicate problems and there can be at most 10 problems in each computer. You have to find the number of problems in each of the computers.
Since there can be multiple solutions. Any valid solution will do.
Input:
Input starts with an integer T (≤ 25), denoting the number of test cases.
Each case starts with a line containing an integer n (0 ≤ n ≤ 20) denoting the total number of problems.
Output:
For each case, print the number of problems stored in each computer in a single line. A single space should separate the non-negative integers.
Sample Input
Output for
Sample Input:
3
10
7
7
Sample output:
0(space)10
0(space)7
1(space)6
my code :
#include<stdio.h>
#include <stdlib.h>
int main()
{
int c,sum;
int i,j,mini=0,maxi;
int com1,com2;
do{
scanf("%d",&c);
}while (c>25);
int t[c+1];
for(i=1; i<=c; i++)
{
do{
scanf("%d",&t[i]);
}while (t[i]>20);
}
for(i=1; i<=c; i++)
{
maxi=t[i];
com1=rand() % (maxi - mini + 1) + mini;
com2=t[i]-com1;
printf("%d %d
",com1,com2);
}
return 0;
}
When I submit my code the judge gives wrong answer.
But while I compile the code in CodeBlocks it gives right answer.
I can't understand the problem in online judge.
How can I solve it.
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