BZOJ 1025 游戏(分组背包)

题目所谓的序列长度实际上就是各循环节的lcm+1.

所以题目等价于求出 一串数之和等于n，这串数的lcm种数。

由唯一分解定理可以联想到只要把每个素数的幂次放在一个分组里，然后对整体做一遍分组背包就行了。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF (LL)<<

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int res=, flag=;

    char ch;

    if((ch=getchar())=='-') flag=;

    else if(ch>=''&&ch<='') res=ch-'';

    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');

    return flag?-res:res;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

int pri[N];

LL dp[N][N];

void init(int n)

{

    mem(pri,);

    FOR(i,,n) {

        if (!pri[i]) pri[++pri[]]=i;

        for (int j=; j<=pri[]&&pri[j]<=n/i; ++j) {

            pri[pri[j]*i]=;

            if (i%pri[j]==) break;

        }

    }

}

int main ()

{

    int n;

    scanf("%d",&n);

    init(n);

    FOR(i,,n) dp[][i]=;

    FOR(i,,pri[]) FOR(j,,n) {

        dp[i][j]=dp[i-][j];

        for (int k=pri[i]; k<=j; k*=pri[i]) dp[i][j]+=dp[i-][j-k];

    }

    printf("%lld\n",dp[pri[]][n]);

    return ;

}