B - Equal Sum Sets
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Let us consider sets of positive integers less than or equal to n. Note that all elements of a set aredifferent. Also note that the order of elements doesnt matter, that is, both {3, 5, 9} and {5, 9, 3} meanthe same set.Specifying the number of set elements and their sum to be k and s, respectively, sets satisfying theconditions are limited. When n = 9, k = 3 and s = 23, {6, 8, 9} is the only such set. There may bemore 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.InputThe 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 assume1 ≤ n ≤ 20, 1 ≤ k ≤ 10 and 1 ≤ s ≤ 155.The end of the input is indicated by a line containing three zeros.OutputThe output for each dataset should be a line containing a single integer that gives the number of thesets 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
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{
int n,k,s;
while(~scanf("%d%d%d",&n,&k,&s))
{
if(k+s+n==0)
break;
int sum1=0,i,j;
for(i=2;i<(1<<(n+1));i++)
{
if(i%2==0)
continue;
int sum=0,ans=0;
for(j=1;j<=n;j++)
if(i&(1<<j))
{
sum+=j;ans++;
}
if(sum==s&&ans==k)
sum1++;
}
printf("%d\n",sum1);
}
}
#include<iostream>
#include<cstdio>
using namespace std;
int n,k,s;
int ans;
int dfs(int nn,int kk,int ss)
{
if(kk==k&&ss==s)
{
ans++;
return ans;
}
for(int i=nn-1;i>=1;i--)
{
if(ss+i<=s)
dfs(i,kk+1,ss+i);
}
}
int main()
{
while(~scanf("%d%d%d",&n,&k,&s))
{
if(s+k+n==0)
break;
ans=0;
dfs(n+1,0,0);
printf("%d\n",ans);
}
return 0;
}
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int n,k,s;
int i,j;
int dp[156][21][11];
int main()
{
memset(dp,0,sizeof(dp));
for(i=1;i<=155;i++)
for(j=1;j<=20;j++)
for(k=1;k<=10;k++)
{
if(k==1)
{
if(i<=j)
dp[i][j][k]=1;
}
else
{
dp[i][j][k]=dp[i][j-1][k];
if(i>j)
dp[i][j][k]+=dp[i-j][j-1][k-1];
}
}
while(cin>>n>>k>>s)
{
if(n+k+s==0)
break;
cout<<dp[s][n][k]<<endl;
}
return 0;
}
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