Sumsets

HDU 2709Problem DescriptionFarmer 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+12) 1+1+1+1+1+23) 1+1+1+2+24) 1+1+1+45) 1+2+2+26) 1+2+4Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000). InputA single line with a single integer, N. OutputThe 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 SourceUSACO 2005 January Silver思路:这题没什么思路,加上数学又学的不好.遂先打表找规律.(1-16)code:#include <iostream>#include<cstdio>#include<cmath>#include<cstring>#define M 100using namespace std;int ans[M];int ant;bool vis[10][10][10][10]; // 1 2 4 8void dfs(int x,int k,int n){if(x==0){int a=0,b=0,c=0,d=0;for(int j=0; j<k; ++j){if(ans[j]==1)a++;if(ans[j]==2)b++;if(ans[j]==4)c++;if(ans[j]==8)d++;}if(!vis[a][b][c][d]){vis[a][b][c][d]=true;for(int j=0; j<k; ++j)printf("%d ",ans[j]);printf("\n");ant++;}return ;}for(int i=1; i<=n; i*=2){if(x-i>=0){ans[k]=i;dfs(x-i,k+1,n);}}}int main(){freopen("xx.out","w",stdout);for(int i=1;i<=16;++i){ant=0;cout<<"n="<<i<<endl;memset(vis,0,sizeof(vis));memset(ans,0,sizeof(ans));dfs(i,0,i);cout<<"Result="<<ant<<endl;}// cout << "Hello world!" << endl;return 0;}

打表结果:

n=1
1 
Result=1
n=2
1 1 
2 
Result=2
n=3
1 1 1 
1 2 
Result=2
n=4
1 1 1 1 
1 1 2 
2 2 
4 
Result=4
n=5
1 1 1 1 1 
1 1 1 2 
1 2 2 
1 4 
Result=4
n=6
1 1 1 1 1 1 
1 1 1 1 2 
1 1 2 2 
1 1 4 
2 2 2 
2 4 
Result=6
n=7
1 1 1 1 1 1 1 
1 1 1 1 1 2 
1 1 1 2 2 
1 1 1 4 
1 2 2 2 
1 2 4 
Result=6
n=8
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 2 
1 1 1 1 2 2 
1 1 1 1 4 
1 1 2 2 2 
1 1 2 4 
2 2 2 2 
2 2 4 
4 4 
8 
Result=10
n=9
1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 2 
1 1 1 1 1 2 2 
1 1 1 1 1 4 
1 1 1 2 2 2 
1 1 1 2 4 
1 2 2 2 2 
1 2 2 4 
1 4 4 
1 8 
Result=10
n=10
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 2 
1 1 1 1 1 1 2 2 
1 1 1 1 1 1 4 
1 1 1 1 2 2 2 
1 1 1 1 2 4 
1 1 2 2 2 2 
1 1 2 2 4 
1 1 4 4 
1 1 8 
2 2 2 2 2 
2 2 2 4 
2 4 4 
2 8 
Result=14
n=11
1 1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 2 
1 1 1 1 1 1 1 2 2 
1 1 1 1 1 1 1 4 
1 1 1 1 1 2 2 2 
1 1 1 1 1 2 4 
1 1 1 2 2 2 2 
1 1 1 2 2 4 
1 1 1 4 4 
1 1 1 8 
1 2 2 2 2 2 
1 2 2 2 4 
1 2 4 4 
1 2 8 
Result=14
n=12
1 1 1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 2 
1 1 1 1 1 1 1 1 2 2 
1 1 1 1 1 1 1 1 4 
1 1 1 1 1 1 2 2 2 
1 1 1 1 1 1 2 4 
1 1 1 1 2 2 2 2 
1 1 1 1 2 2 4 
1 1 1 1 4 4 
1 1 1 1 8 
1 1 2 2 2 2 2 
1 1 2 2 2 4 
1 1 2 4 4 
1 1 2 8 
2 2 2 2 2 2 
2 2 2 2 4 
2 2 4 4 
2 2 8 
4 4 4 
4 8 
Result=20
n=13
1 1 1 1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 1 2 
1 1 1 1 1 1 1 1 1 2 2 
1 1 1 1 1 1 1 1 1 4 
1 1 1 1 1 1 1 2 2 2 
1 1 1 1 1 1 1 2 4 
1 1 1 1 1 2 2 2 2 
1 1 1 1 1 2 2 4 
1 1 1 1 1 4 4 
1 1 1 1 1 8 
1 1 1 2 2 2 2 2 
1 1 1 2 2 2 4 
1 1 1 2 4 4 
1 1 1 2 8 
1 2 2 2 2 2 2 
1 2 2 2 2 4 
1 2 2 4 4 
1 2 2 8 
1 4 4 4 
1 4 8 
Result=20
n=14
1 1 1 1 1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 1 1 2 
1 1 1 1 1 1 1 1 1 1 2 2 
1 1 1 1 1 1 1 1 1 1 4 
1 1 1 1 1 1 1 1 2 2 2 
1 1 1 1 1 1 1 1 2 4 
1 1 1 1 1 1 2 2 2 2 
1 1 1 1 1 1 2 2 4 
1 1 1 1 1 1 4 4 
1 1 1 1 1 1 8 
1 1 1 1 2 2 2 2 2 
1 1 1 1 2 2 2 4 
1 1 1 1 2 4 4 
1 1 1 1 2 8 
1 1 2 2 2 2 2 2 
1 1 2 2 2 2 4 
1 1 2 2 4 4 
1 1 2 2 8 
1 1 4 4 4 
1 1 4 8 
2 2 2 2 2 2 2 
2 2 2 2 2 4 
2 2 2 4 4 
2 2 2 8 
2 4 4 4 
2 4 8 
Result=26
n=15
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 1 1 1 2 
1 1 1 1 1 1 1 1 1 1 1 2 2 
1 1 1 1 1 1 1 1 1 1 1 4 
1 1 1 1 1 1 1 1 1 2 2 2 
1 1 1 1 1 1 1 1 1 2 4 
1 1 1 1 1 1 1 2 2 2 2 
1 1 1 1 1 1 1 2 2 4 
1 1 1 1 1 1 1 4 4 
1 1 1 1 1 1 1 8 
1 1 1 1 1 2 2 2 2 2 
1 1 1 1 1 2 2 2 4 
1 1 1 1 1 2 4 4 
1 1 1 1 1 2 8 
1 1 1 2 2 2 2 2 2 
1 1 1 2 2 2 2 4 
1 1 1 2 2 4 4 
1 1 1 2 2 8 
1 1 1 4 4 4 
1 1 1 4 8 
1 2 2 2 2 2 2 2 
1 2 2 2 2 2 4 
1 2 2 2 4 4 
1 2 2 2 8 
1 2 4 4 4 
1 2 4 8 
Result=26
n=16
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 
1 1 1 1 1 1 1 1 1 1 1 1 2 2 
1 1 1 1 1 1 1 1 1 1 1 1 4 
1 1 1 1 1 1 1 1 1 1 2 2 2 
1 1 1 1 1 1 1 1 1 1 2 4 
1 1 1 1 1 1 1 1 2 2 2 2 
1 1 1 1 1 1 1 1 2 2 4 
1 1 1 1 1 1 1 1 4 4 
1 1 1 1 1 1 1 1 8 
1 1 1 1 1 1 2 2 2 2 2 
1 1 1 1 1 1 2 2 2 4 
1 1 1 1 1 1 2 4 4 
1 1 1 1 1 1 2 8 
1 1 1 1 2 2 2 2 2 2 
1 1 1 1 2 2 2 2 4 
1 1 1 1 2 2 4 4 
1 1 1 1 2 2 8 
1 1 1 1 4 4 4 
1 1 1 1 4 8 
1 1 2 2 2 2 2 2 2 
1 1 2 2 2 2 2 4 
1 1 2 2 2 4 4 
1 1 2 2 2 8 
1 1 2 4 4 4 
1 1 2 4 8 
2 2 2 2 2 2 2 2 
2 2 2 2 2 2 4 
2 2 2 2 4 4 
2 2 2 2 8 
2 2 4 4 4 
2 2 4 8 
4 4 4 4 
4 4 8 
8 8 
16 
Result=36

可以发现规律了: 当n是奇数的时候,值就等于前面那个偶数的值.n为偶数的时候，它的值等于前面那个奇数的值再加上它本身/2的那个值AC code:#include<iostream>#include<cstdio>using namespace std;#define M 1000000typedef long long ll;const int mode=1000000000;int n;ll num[M]={0,1,2};void slove(){int i,j;for(i=3;i<=M;++i){if(i&1) num[i]=num[i-1]; //奇数就等于前一位的偶数else num[i]=(num[i-1]+num[i/2])%mode;//偶数等于前一位+当前数/2}}int main(){slove();while(scanf("%d",&n)!=EOF){//printf("%I64d\n",num);printf("%I64d\n",num%mode);//cout<<num<<endl;}return 0;}