poj 2229 Sumsets 【完全背包 or 递推】
2015-12-20 12:03
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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).
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
Sample Output
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。。。
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]。
Time Limit: 2000MS | Memory Limit: 200000K | |
Total Submissions: 14968 | Accepted: 5978 |
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|1 using 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 ); 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|1 using 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 ); return 0; }
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