poj 2229 Sumsets 【完全背包 or 递推】

Sumsets

Time Limit: 2000MS Memory Limit: 200000KTotal Submissions: 14968 Accepted: 5978

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

7

Sample Output

6

题意：给定一个N，只允许使用2的幂次数，问有多少种不同的方案组成N。

首先这道题有一个很好想的思路：裸完全背包

最多20个物品，价值分别为2^0, 2^1, ..., 2^19。

这样直接跑一次完全背包打好表就可以了。 G++可以过，C++过不了。。。

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#define INF 0x3f3f3f#define eps 1e-8#define MAXN (1000000+1)#define MAXM (100000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;LL dp[MAXN];int fac[21];int Pow(int a, int n){    int ans = 1;    while(n)    {        ans *= a;        n--;    }    return ans;}void getdp(){    for(int i = 0; i < 20; i++)        fac[i] = Pow(2, i);    dp[0] = 1;    for(int i = 0; i < 20; i++)        for(int j = fac[i]; j < MAXN; j++)            dp[j] = (dp[j] + dp[j-fac[i]])%1000000000;}int main(){    getdp();    int n; Ri(n); Pl(dp[n]);    return 0;}

另外一种思路：

两种dp状态推导dp[]。

一、dp[i] = dp[i-1];

二、dp[i] = dp[i-1] + dp[i>>1]。

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#define INF 0x3f3f3f#define eps 1e-8#define MAXN (1000000+1)#define MAXM (100000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1using namespace std;LL dp[MAXN];void getdp(){    dp[0] = 1;    for(int i = 1; i < MAXN; i++)    {        if(i & 1)            dp[i] = dp[i-1];        else            dp[i] = (dp[i-1] + dp[i >> 1]) % 1000000000;    }}int main(){    getdp();    int n; Ri(n); Pl(dp[n]);    return 0;}