HDU 3092 Least common multiple 01背包

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=3092

Least common multiple

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
#### 问题描述
> Partychen like to do mathematical problems. One day, when he was doing on a least common multiple(LCM) problem, he suddenly thought of a very interesting question: if given a number of S, and we divided S into some numbers , then what is the largest LCM of these numbers? partychen thought this problems for a long time but with no result, so he turned to you for help!
> Since the answer can very big,you should give the answer modulo M.
#### 输入
> There are many groups of test case.On each test case only two integers S( 0 Output the largest LCM modulo M of given S.
####样例输入
> 6 23

样例输出

6

题意

给你一个n，求把n拆成若干个数的和，要求这些数的最小公倍数最大。最后输出%M.

题解

由于最大公倍数会爆，无法直接维护，所以我们可以通过取对数的方式，得到大小，确定转移方向，然后利用这个去维护最优解。

这道题需要用到两个贪心策略：

1、只选质数（或质数的幂）（2^{2*3=12,还不如2}2+3=7。）

2、每个质数只选一个（如果你拆出39=3+9+27,明显，只有27是有用的。。）

然后就可以转换成01背包求解了（贪心策略2决定了这是01背包！不是完全背包）

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<sstream>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;
const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);
//start----------------------------------------------------------------------
const int maxn=3333;
double dp[maxn];
LL ans[maxn];
bool pri[maxn];
VI arr;
void pre() {
    clr(pri,0);
    for(int i=2; i*i<maxn; i++) {
        if(!pri[i]) {
            for(int j=i*i; j<maxn; j+=i) {
                pri[j]=true;
            }
        }
    }
    for(int i=2; i<maxn; i++) if(!pri[i]) arr.pb(i);
}
int main() {
    pre();
    int S,M;
    while(scf("%d%d",&S,&M)==2) {
        clr(dp,0);
        double Ma=-1;
        LL last=1;
        for(int i=0; i<=S; i++) ans[i]=1;
        //01背包
        for(int i=0; arr[i]<=S; i++) {
            for(int j=S; j>=arr[i]; j--) {
                for(int k=arr[i]; k<=j; k*=arr[i]) {
                    if(dp[j]<dp[j-k]+log10(k)) {
                        dp[j]=dp[j-k]+log10(k);
                        ans[j]=ans[j-k]*k%M;
                    }
                }
                if(Ma<dp[j]+eps) {
                    Ma=dp[j];
                    last=ans[j];
                }
            }
        }
        prf("%I64d\n",last);
    }
    return 0;
}
//end-----------------------------------------------------------------------