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
20
1542
5448
1
题目大意：
从 1到n中选k个数 让k个数的和为s;
其实就是暴力枚举
把所有的情况算出来，但是先在题目的数据还是比较大的
所以还要用到状态压缩；
#include<iostream>
using namespace std;
int a[];
int A[];
int ans=;
void dfs(int n,int cur){
if(cur==a[]){
int sum=;
for(int i=;i<a[];i++){
sum+=A[i];
}
if(sum==a[])ans++;
}
int s=;
if(cur!=)s=A[cur-]+;
for(int i=s;i<=n;i++){
A[cur]=i;
dfs(n,cur+);
}
}
int main(){
while(cin>>a[]>>a[]>>a[]&&a[]+a[]+a[]){
ans=;
dfs(a[],);
cout<<ans<<endl;
}
return ;
}