B - Equal Sum Sets

最新推荐文章于 2018-05-03 20:34:51 发布

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本文介绍了一种计算在给定总数和元素数量限制下，由正整数组成的独特集合的数量的方法。通过递归算法和动态规划技术，解决了如何找出所有可能的组合而不重复计数的问题。

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

1542

5448

#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;
}