POJ 2229 Sumsets
Description
Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer power of 2. Here are the possible sets of numbers that sum to 7:
1) 1+1+1+1+1+1+1
2) 1+1+1+1+1+2
3) 1+1+1+2+2
4) 1+1+1+4
5) 1+2+2+2
6) 1+2+4
Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000).
1) 1+1+1+1+1+1+1
2) 1+1+1+1+1+2
3) 1+1+1+2+2
4) 1+1+1+4
5) 1+2+2+2
6) 1+2+4
Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000).
Input
A single line with a single integer, N.
Output
The number of ways to represent N as the indicated sum. Due to the potential huge size of this number, print only last 9 digits (in base 10 representation).
Sample Input
7
Sample Output
6
Source
这题是看一个数能用几个不同二次方幂的数和表示
首先来练习一下深搜把,这一开始我没想到还能用深搜,尽管不对,当然可慢啦啦啦啦
#include<stdio.h> #include<math.h> #include<iostream> using namespace std; int ans; long long p[50]; void dfs(int n,int last) { if(n==0) { ans++; return; } for(int i=0;n-p[i]>=0;i++) { if(p[i]>=last) dfs(n-p[i],p[i]); } } int main() { int n; for(int i=0;i<=49;i++) p[i]=pow(2,i); cin>>n; dfs(n,0); printf("%d",ans); return 0; }
接下来是dp版的,不知道为啥,还是超时!!!!,但思想要学习下,看懂下面的那个dp更新表就欧克了
dp表假设只有1,然后添加2,然后添加4然后添加8,更新表
#include<iostream> #include<string.h> #include<stdio.h> using namespace std; long long dp[1000001],p[30]; int main() { int n; while(scanf("%d",&n)!=EOF) { memset(dp,0,sizeof(dp)); p[0]=dp[0]=1; for(int i=1;i<=29;i++) { p[i]=p[i-1]<<1; } for(int i=0;i<=29;i++) { if(p[i]<=n) { for(int j=p[i];j<=n;j++) { dp[j]=(dp[j]+dp[j-p[i]])%1000000000; } } } /*dp 1 2 3 4 5 6 7//假设输入7 1 1 1 1 1 1 1 1 2 2 2 3 3 4 4 4 4 4 6 6 */ cout<<dp[n]<<endl; } }
最后过的是这个,我他喵。。。
#include<iostream> #include<string.h> using namespace std; long long a[1000001]; int main() { int n; a[1]=1,a[2]=2; for(int i=3;i<1000001;i++) { a[i]=a[i-2]+a[i/2]; a[i]%=1000000000; } cin>>n; cout<<a[n]; }
1.n为奇数,a[n]=a[n-1]
2.n为偶数:
(1)如果加数里含1,则一定至少有两个1,即对n-2的每一个加数式后面 +1+1,总类数为a[n-2];
(2)如果加数里没有1,即对n/2的每一个加数式乘以2,总类数为a[n/2];
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