D.6661 - Equal Sum Sets

Equal Sum Sets

Let us consider sets of positive integers less than or equal to n. Note that all elements of a set are
different. Also note that the order of elements doesnt matter, that is, both {3, 5, 9} and {5, 9, 3} mean
the same set.
Specifying the number of set elements and their sum to be k and s, respectively, sets satisfying the
conditions are limited. When n = 9, k = 3 and s = 23, {6, 8, 9} is the only such set. There may be
more than one such set, in general, however. When n = 9, k = 3 and s = 22, both {5, 8, 9} and {6, 7, 9}
are possible.
You have to write a program that calculates the number of the sets that satisfy the given conditions.
Input
The input consists of multiple datasets. The number of datasets does not exceed 100.
Each of the datasets has three integers n, k and s in one line, separated by a space. You may assume
1 ≤ n ≤ 20, 1 ≤ k ≤ 10 and 1 ≤ s ≤ 155.
The end of the input is indicated by a line containing three zeros.
Output
The output for each dataset should be a line containing a single integer that gives the number of the
sets that satisfy the conditions. No other characters should appear in the output.
You can assume that the number of sets does not exceed 231 − 1.

Sample Input
9 3 23
9 3 22
10 3 28
16 10 107
20 8 102
20 10 105
20 10 155
3 4 3
4 2 11
0 0 0
Sample Output
1
2
0
20
1542
5448
1
0
0

题意：

在1~n 中任意选 k 个数组成 s 。求共有多少种组法。

分析：

枚举。普通枚举会超时，N!次

因为n,k,s的范围很小，可以直接用dfs。

代码：

 #include<cstdio>
 #include<iostream>
 using namespace std;

 int n,k,s;
 int ans;

 void dfs(int num,int kase,int sum)
 {
     if(kase==k&&sum==s)
     {
         ans++;
         return;
     }
     if(kase>k||sum>s)
         return;
     else
         for(int i=num+;i<=n;i++)
             dfs(i,kase+,sum+i);
             return;
 }

 int main ()
 {
     while(scanf("%d%d%d",&n,&k,&s)!=EOF)
     {
         if(n==&&k==&&s==)
             break;
         ans=;
         for(int i=;i<=n;i++)
             dfs(i,,i);
         printf("%d\n",ans);
     }
     return ;
 }

这道题还可以用dp 方法来做，但是还没有学。学了之后再重新补上。