poj2229(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
设dp[n]表示n的拆分数，若n是奇数，必能拆分出一个1，所以dp[n]=dp[n-1]；若n是偶数，则n/2的所有拆分方法每个数乘以2后就是n的拆分方法，n还可以拆成1+1+(n-2)，所以dp[n]=dp[n/2]+dp[n-2]
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double EPS = 1e-8;
const int MAXN = 1000010;
int dx[] = {0, 1, 0, -1}, dy[] = {-1, 0, 1, 0};
int dp[MAXN] = {0, 1, 2};
int main()
{
for (int i = 3; i < MAXN; ++i)
dp[i] = (i&1) ? dp[i-1] : (dp[i/2] + dp[i-2]) % 1000000000;
int n;
cin >> n;
cout << dp[n] << endl;
return 0;
}算法码上来 文章被收录于专栏
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