1001. Black Magic (soj)

http://soj.me/show_problem.php?pid=1001&cid=1004

1001. Black Magic

Description

Believe it or not, there is a powerful black magic, invented by Mage Philia, which will make one person fall in love with an OTAKU. Needless to say, this is the ultimate goal for all of you. However, in order to learn this spell, you need to solve a problem first.

First, Mage Philia wrote a row of numbers from 1…N on the blackboard. Then, she copied the last row of numbers to the next line, and erased the first M number in the new row. She repeated the same action again and again until all the number in the new row had been erased. For example, let N = 5 and M = 2, all the numbers on the blackboard are as follows.

1	2	3	4	5
		3	4	5
				5

At last, Mage Philia generates a sequence by concatenating all the rows from top to bottom. In the above example, the sequence is [1, 2, 3, 4, 5, 3, 4, 5, 5]

Here comes your problem, find out the K^th number in the sequence.

Input

There are multiple test cases.

The first line contains an integer T (1 ≤ T ≤ 20000), indicating the number of test cases.

T lines follows. Each line contains three integers, N, M, K (1 ≤ N, M, K ≤ 10000), whose meanings are described above.

Output

Output one line for each test case. In each line, there is a number indicating the K^th number in the sequence. If K is greater than the length of the sequence, output a single 0 instead.

Sample Input

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3
5 2 1
5 2 7
5 2 9

Sample Output

1
4
5

分析：模拟暴力即可

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<string.h>
#include<string>
#include<ctime>
#include<queue>
#include<list>
#include<map>
#include<set>
#include<vector>
#include<stack>
using namespace std;
int main()
{
    int t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            int n,m,k;
            scanf("%d%d%d",&n,&m,&k);
            int s=k,left=n,temp=0;
            while(1)
            {
                if(s>left)
                    s-=left;
                else
                    break;
                temp+=m;
                left-=m;
                if(left<=0)
                    break;
            }
            if(s+temp>n)
                printf("0\n");
            else
                printf("%d\n",s+temp);

        }
    }
    return 0;
}