POJ2229 Sumsets
2017-09-05 07:53
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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).
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
7Sample Output
6找规律做出
多写几组数据:
1 -> 1
2 -> 2
3 -> 2
4 -> 4
5 -> 4
6 -> 6
7 -> 6
就可以看出来当n为奇数,ans=sum[n-1],n为偶数时,ans=sum[n-1]+sum[n/2]
CODE
#include <stdio.h> int sum[1000000]; int main() { int n; while(scanf("%d",&n)!=EOF) { sum[1]=1; sum[2]=2; for(int i=3;i<=n;i++) { if(i%2!=0) sum[i]=sum[i-1]; else sum[i]=(sum[i-2]+sum[i/2])%1000000000; } printf("%d\n",sum ); } }
看了题解后,这不是规律,这是有道理的!而且这题是可以用动态规划做的!
所谓找规律的道理所在:
设a为和为 n 的种类数;
根据题目可知,加数为2的N次方,即 n 为奇数时等于它前一个数 n-1 的种类数 a[n-1] ,若 n 为偶数时分加数中有无 1 讨论,即关键是对 n 为偶数时进行讨论:
1.n为奇数,a
=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];
所以总的种类数为:a
=a[n-2]+a[n/2];
动态规划思想:
附一个完全背包解法,暂时没有想出这个代码的含义#include <iostream> #include <cstdio> #include <string.h> #include <stack> #include <queue> #include <map> #include <set> #include <vector> #include <math.h> #include <algorithm> using namespace std; #define LL long long #define INF 0x3f3f3f3f const double pi = acos(-1.0); const int mod = 1e9; const int N =1e6+10; int dp ; void Init() { memset(dp,0,sizeof(dp)); dp[0]=dp[1]=1; for(int i=1;i<=22;i++) { for(int j=2;j<=1000000;j++) { int v=1<<(i-1); if(j>=v) dp[j]=dp[j-v]+dp[j]; while(dp[j]>mod) dp[j]-=mod; } } } int main() { int n; Init(); while(~scanf("%d",&n)) printf("%d\n",dp ); return 0; }
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