POJ 2229 Sumsets(递推,思考)

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 
 5 #include<string> 
 6 #include<vector>
 7 #include<set>
 8 #include<map>
 9 #include<queue>
10 #include<math.h>
11 
12 #include<algorithm>
13 #include<iostream>
14 
15 const int INF = 0x7f7f7f7f;
16 using namespace std;
17 typedef long long ll;
18 
19 int dp[1000005];
20 int main(){
21 
22     /*
23         若i为偶数 : 若有  1 ,至少有两个    ---->f[i-2]的情况+两个1, 
24                     若没有1 , 将偶数分解/2-----> f[i/2];
25           则dp[i] = dp[i / 2] + dp[i-2]
26            
27           若i为奇数: 必定有 1, 
28             dp[i] = 减去1的序列,加上1 
29           则dp[i] = dp[i – 1]   
30     */
31     int n;
32     dp[0] = 0;
33     dp[1] = 1;
34     dp[2] = 2;
35     for(int i = 3 ; i <= 1000000 ; i ++){
36         if(i&1) dp[i] = dp[i-1]%1000000000;
37         else dp[i] = (dp[i-2]+dp[i/2])%1000000000;
38     }
39     while(scanf("%d",&n) != EOF && n){
40         printf("%d\n",dp[n]%1000000000);
41     }
42     
43     return 0;
44 }