Poj 2229（dp）

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 10974		Accepted: 4419

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

Source

USACO 2005 January Silve r

这题我开始的想法是当成一个完全背包做，logn种物品，所以复杂度nlogn。结果无限TLE，无语了。然后搜题解，看到了O（n）的做法，也挺巧妙的。当n为奇数时，dp[n] = dp[n-1],因为一定包含一个1，所以dp[n-1]的每一种再加1都一一对应着dp[n];当n为偶数时，要么至少含2个1，或者不含1，不含1时，方法数就是取dp[n/2]的每一种物品乘以2价值的物品。所以dp[n] = dp[n-2] + dp[n/2]。最后说明nlogn的解法也是能过的，要把mod运算改为减法，我刚好卡2s过。

O(nlogn)

int dp[25][maxn];

void solve(){
    memset(dp,0,sizeof(dp));
    for(int i = 0;i < 25;i++) dp[i][1] = dp[i][0] = 1;
    for(int i = 1;i < 25;i++){
        for(int j = 2;j < maxn;j++){
            int v = 1 << (i-1);
            if(j >= v) dp[i][j] = (LL(dp[i][j-v]) + dp[i-1][j]);
            else dp[i][j] = dp[i-1][j];
            while(dp[i][j] > Mod) dp[i][j] -= Mod;
        }
    }
}

int main(){
    int n;
    solve();
    while(cin >> n){
        cout << dp[24][n] << endl;
    }
    return 0;
}

O(n)

LL dp[maxn];

void solve(){
    dp[0] = 1;
    for(int i = 1;i < maxn;i++){
        if(i%2 == 1) dp[i] = dp[i-1];
        else dp[i] = (dp[i-2] + dp[i/2]) % Mod;
    }
}

int main(){
    int n;
    solve();
    while(scanf("%d",&n) != EOF){
        printf("%d\n",dp[n]);
    }
    return 0;
}