POJ 2229 Sumsets(基础dp)
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Sumsets
Time Limit: 2000MS Memory Limit: 200000KTotal Submissions: 20137 Accepted: 7870
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
AC代码:
思路:i是奇数 dp[i] = dp[i-1], i是偶数 dp[i] = dp[i-1]+dp[i/2],也可以算一道递推的题目吧
#include <iostream> #include <cstdio> using namespace std; #define maxn 1000005 #define INF 1000000000 int dp[maxn]; void solve() { dp[0] = 1; for(int i = 1; i < maxn; i++) { if(i % 2 != 0) dp[i] = dp[i-1]; // else dp[i] = ( dp[i-1] + dp[i/2] ) % INF; } } int main() { int n; solve(); while(scanf("%d" , &n) != EOF) { printf("%d\n" , dp[n]); } return 0; }
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