SDUT3143:Combinatorial mathematics(组合数学）

题意：传送门

题目描述

As you know, shadow95 is pretty good at maths, especially combinatorial mathematics. Now, he has made a problem for you. We call a subset which exactly has r elements as a "r-subset".For example, {1,2,5} is a 3-subset(r=3) of {1,2,3,4,5}. Now, your task is to form all the r-subset of {1,2,...,n}, then output them in lexicographic order（字典序）.

输入

 The input file ends by EOF.

For each test case, there are two integer n,r.(1 <= r < n <= 20)

输出

 First output the case number, then output all the r-subset of {1,2,...,n} in lexicographic order.

Each case seperates by a blank line.

示例输入

3 2
3 1

示例输出

Case #1:
1 2
1 3
2 3

Case #2:
1
2
3
题目很简单，代码很快就敲出来了，令人伤心的是PE无数遍，因为之前的题都没卡那么严，遇到这种输出格式错误慌了.......
每组数据之后有一个空行，但是最后一组数据没有空行，如果只输入一组n,m，那么之后不需要打印空行。
具体请看代码：

#include <iostream>
#include <algorithm>
#include <math.h>
#include <map>
#include <queue>
#include <stack>
#define inf 0x3f3f3f3f
#include <stdio.h>
#include <string.h>
typedef long long ll;
#define mod 10000007
#define eps 1e-9
using namespace std;
int n,r,a[];
void dfs(int k,int step)
{
    a[step]=k;
    if(step>r) return ;
    if(step==r)
    {
        for(int i=; i<=step; i++)
        {
            if(i==) printf("%d",a[i]);
            else printf(" %d",a[i]);
        }
        printf("\n");
    }
    for(int i=k+; i<=n; i++)
        dfs(i,step+);
}
int main()
{
    int K=;
    while(scanf("%d%d",&n,&r)!=EOF)
    {
        if(K>=1) printf("\n");//这样能避免PE
        printf("Case #%d:\n",++K);
        for(int i=; i<=n-r+; i++)
            dfs(i,);
    }
    return ;
}