SGU 116 Index of super-prime 数论+背包_116. index of super-prime time limit per test: 0.2-优快云博客

本文介绍了一种结合数论和背包动态规划的问题解决方法。通过预先计算素数并使用动态规划来找出特定数值的分解方式，该算法有效地解决了问题，并能够输出具体的分解路径。

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不错的题目！数论+背包dp，dp的时候附加计入一个个数就可以了~

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;

#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif

#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a)        for( int i = (a)-1 ; i >= 0 ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define SZ(a)           ((int)a.size())
#define PP(n,m,a)       puts("---");FF(i,n){FF(j,m)cout << a[i][j] << ' ';puts("");}
#define pb              push_back
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-10;
const double pi = acos(-1.0);
const int maxn = 10011;

vector<int>p;
vector<int>vp;
bool isp[maxn];

int div(int x)
{
    int temp = sqrt(x);
    for(int i=1;i<p.size();i++)
    {
        if(x%p[i]==0)
        {
            return true;
        }
        if(p[i]>temp)
        {
            return false;
        }
    }
    return false;
}

void init()
{
    vp.clear();
    p.clear();
    MM(isp,false);
    p.push_back(0);
    p.push_back(2);
    isp[2]=true;
    for(int i=3;i<maxn;i++)
    {
        if(!div(i))
        {
            isp[i]=true;
            p.push_back(i);
        }
    }
    for(int i=1;i<p.size();i++)
    {
        if(isp[i])
        {
            vp.push_back(p[i]);
        }
    }
    return ;
}

bool can[maxn];
int back[maxn];
int re[maxn];


void dpstart()
{
    MM(can,false);
    can[0]=true;
    MM(back,0);
    for(int i=0;i<maxn;i++)
    {
        re[i]=inf;
    }
    re[0]=0;
    int now,temp;
    for(int i=0;i<vp.size();i++)
    {
        now = vp[i];
        for(int j=1;j<maxn;j++)
        {
            if(j-now>=0)
            temp = re[j-now] + 1;
            if( j-now>=0 && can[j-now] && temp < re[j])
            {
                can[j] = true;
                back[j] = now;
                re[j] = temp;
            }
        }
    }
    return ;
}

int n;
vector<int>v;

int main()
{
    init();
    dpstart();


    cin>>n;
    v.clear();
    if(can[n])
    {
        int temp = n;
        int now;
        while(back[temp])
        {
            now = back[temp];
            v.pb(now);
            temp -= back[temp];
        }
        cout<<v.size()<<endl;
        for(int i=0;i<v.size()-1;i++)
        {
            cout<<v[i]<<" ";
        }
        cout<<v.back()<<endl;
    }
    else
    {
        cout<<"0"<<endl;
    }
    return 0;
}