Sumsets
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1977 Accepted Submission(s): 781
Problem 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).
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
dp[n]=dp[n-1]+dp[n-2]+dp[n-4]+dp[n-8].....
防止重複的方法是 假設2最大來一次 4 最大來一....
#include<iostream>
#include<cstring>
#define N 1000005
using namespace std;
long long dp[N]={0};
long long c[25]={0};
int main()
{
int n=0;
c[0]=1;
int t=0;
for(t=1;t<=24;t++)
{
c[t]=c[t-1]*2;
}
dp[0]=1;
for(int i=0;i<24&&c[i]<=N;i++)
{
for(int j=c[i];j<=N;j++)
{
dp[j]=(dp[j]+dp[j-c[i]])%1000000000;
}
}
while(scanf("%d",&n)!=EOF)
{
printf("%lld\n",dp[n]);
}
return 0;
}