URAL 1731. Dill（数学啊 ）

题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1731

1731. Dill

Time limit: 0.5 second

Memory limit: 64 MB

Ivan Vasil'evich likes to work in his garden. Once he heard that dill was a very beautiful and healthy crop and planted his whole garden with two varieties of it.

When the dill was ripe, Ivan Vasil'evich harvested it and put it into boxes. He filled n boxes with dill of the first variety and m boxes with dill of the second variety. The weight
of each box with dill in kilograms was an integer and all the weights were different. In order to please his neighbors Ivan Ivanovich and Ivan Nikiforovich, Ivan Vasil'evich decided to give each of them two boxes with dill, one box of each variety. Ivan Vasil'evich
didn't want Ivan Ivanovich and Ivan Nikiforovich to quarrel, so he decided that the total weight of the boxes given to each of the neighbors should be equal. Ivan Vasil'evich considered all the possible variants and saw that this was impossible. Find the weights
of all the boxes with dill stocked by Ivan Vasil'evich.

Input

The only input line contains space-separated integers n and m (2 ≤ n, m ≤ 50).

Output

In the first line output n space-separated integers. These must be the weights of the boxes with the first variety of dill. In the second line output m space-separated integers, which
are the weights of the boxes with the second variety of dill. The integers must be positive, different, and should not exceed 109. If there are several solutions, output any of them.
If there is no solution, output the line “It is a lie!”.

Sample

input	output
2 3	1 2 5 10 12

代码例如以下：

#include <cstdio>
#include <cstring>
#define LL __int64
int main()
{
    int n, m;

    int c1[57], c2[57];
    while(~scanf("%d%d",&n,&m))
    {
        int a = 1;
        for(int i = 1; i <= n; i++)
        {
            c1[i] = a++;
        }
        c2[1] = a;
        for(int i = 2; i <= m; i++)
        {
            c2[i] = a+c2[i-1];
        }
        printf("%d",c1[1]);
        for(int i = 2; i <= n; i++)
        {
            printf(" %d",c1[i]);
        }
        printf("\n");
        printf("%d",c2[1]);
        for(int i = 2; i <= m; i++)
        {
            printf(" %d",c2[i]);
        }
        printf("\n");
    }
    return 0;
}