Sumsets (Full backpack)

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 15045		Accepted: 5997

Description

Farmer John commanded his cows to search for different sets of numbers the sum to a given number. The cows use is numbers that is an integer power of 2. Here is 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 representation).

Sample Input

7

Sample Output

6
The puzzle: Let the number of 2^k with the sum of N, just see that is the mother function, did not fight out, and certainly not because the amount of data is too large to definitely time out;
Can actually write with a complete backpack;
Thinking very well, the 2^k of the goods to the backpack inside put; dp[j]+=dp[j-i] for the number of programs;
Recursion is also very good to think, if it is odd, direct dp[i]=dp[i-1]
If it is an even number, dp[i] can be obtained by DP[I/2], or it can be obtained by dp[i-1];
Full backpack:

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include < Algorithm>using namespace std, #define MEM (x, y) memset (x,y,sizeof (×)) #define T_T while (t--) #define SI (x) scanf ("%d", &X) #define SL (x) scanf ("%lld", &x) typedef long LONG ll;const int maxn=1000010;const int Mod=1e9;int bag[maxn];int Main () {int N; ~SCANF ("%d", &n) {mem (bag,0); bag[0]=1; for (int. i=1;i<=n;i*=2) {for (int j=i;j<=n;j++) { Bag[j]+=bag[j-i]; Bag[j]%=mod; }} printf ("%d\n", Bag[n]); }return 0;}

The parent function timed out:

#include <stdio.h>const int maxn= 1000010;int main () {    int a[maxn],b[maxn],n;    while (~SCANF ("%d", &n)) {        int i,j,k;        for (i=0;i<=n;i++) {            a[i]=1;b[i]=0;        }        for (i=2;i<=n;i*=2) {for            (j=0;j<=n;j++) for (                k=0;k+j<=n;k+=i)                    b[j+k]+=a[j];                for (j=0;j<=n;j++)                    a[j]=b[j],b[j]=0;            }        printf ("%d\n", A[n]);    }    return 0;}

recursion;

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include < Algorithm>using namespace std, #define MEM (x, y) memset (x,y,sizeof (×)) #define T_T while (t--) #define SI (x) scanf ("%d", &X) #define SL (x) scanf ("%lld", &x) typedef long LONG ll;const int maxn=1000010;const int Mod=1e9;int dp[maxn];int m Ain () {int n;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/2])%MOD;} while (~SCANF ("%d", &n)) printf ("%d\n", Dp[n]); return 0;}

　　

Sumsets (Full backpack)