[动态规划]Tak and Cards
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题目描述
Tak has N cards. On the i-th (1≤i≤N) card is written an integer xi. He is selecting one or more cards from these N cards, so that the average of the integers written on the selected cards is exactly A. In how many ways can he make his selection?
Constraints
1≤N≤50
1≤A≤50
1≤xi≤50
N,A,xi are integers.
Partial Score
200 points will be awarded for passing the test set satisfying 1≤N≤16.
Constraints
1≤N≤50
1≤A≤50
1≤xi≤50
N,A,xi are integers.
Partial Score
200 points will be awarded for passing the test set satisfying 1≤N≤16.
输入
The input is given from Standard Input in the following format:
N A
x1 x2 … xN
N A
x1 x2 … xN
输出
Print the number of ways to select cards such that the average of the written integers is exactly A.
样例输入
4 8
7 9 8 9
样例输出
5
提示
The following are the 5 ways to select cards such that the average is 8:
Select the 3-rd card.
Select the 1-st and 2-nd cards.
Select the 1-st and 4-th cards.
Select the 1-st, 2-nd and 3-rd cards.
Select the 1-st, 3-rd and 4-th cards.
思路:很显然,是要求算,这里dp[i][j]表示取i个值和为j的取法数。C(n,1)+C(n,2)+...+C(n,n)=2^n-1,暴力枚举显然超时;考虑找到dp[i][j]的递推式;
容易想到dp[i][j]+=dp[i-1][j-a[k]](k=1~n),也就是说取i个值和为j的方法数等于取i-1个值和为j-a[k](k=1~n)的方法数之和?显然是不对的,拿样例举例,计算dp[3][24](初值为0)时,在k=1时,dp[3][24]+=dp[2][17],k=2时,dp[3][24]+=dp[2][15],k=3时,dp[3][24]+=dp[2][16]...暂且先看k=1~3这一部分,并暂且认为已计算出dp[2][j]的所有结果都是正确的,那dp[3][24]在被k=1~3更新时,很显然能看见取前三个数这种取法被重复计算了三次(也就是这一种取法被视为了三种取法),那如何避免这样的重复计算呢——在计算dp[i][j]考虑取a[k]时,取满j-a[k]的另外i-1个值只能从a[1~k]之中取。这样的话,就变成了考虑用a[k]更新所有的dp[i][j](i=k~1,j=sum[k]~a[k])(注意用a[k]更新dp[i][j]的顺序——更新dp[i][j]要用到dp[i-1][j-a[k]],不能用a[k]先把dp[i-1][j-a[k]]给更新了),而不是考虑将dp[i][j]用所有的a[k](k=1~n)去更新;
AC代码:
#include <iostream> #include<cstdio> typedef long long ll; using namespace std; int a[55]; int sum[55]; ll dp[55][2505]; int main() { int n,A; scanf("%d%d",&n,&A); for(int i=1;i<=n;i++) {scanf("%d",&a[i]); sum[i]=sum[i-1]+a[i];} dp[0][0]=1; for(int k=1;k<=n;k++){ for(int i=k;i>=1;i--){ for(int j=sum[k];j>=a[k];j--){ dp[i][j]+=dp[i-1][j-a[k]]; } } } ll ans=0; for(int i=1;i<=n;i++) ans+=dp[i][i*A]; printf("%lld\\n",ans); return 0; }
另外,关于动态规划和递推,我已经分不清了(#°Д°)
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