1103. Integer Factorization (30)

1103. Integer Factorization (30)
分析：显然有明显的递归特征：计算n,k,p，实际上只需求出n-i^p,k-1,p；考察回溯；
思路：每次保存该条路径，直至当前情况的递归终止，递归终止时，判断是否满足题设条件；用vmax记录最大因子序列和值，cmax记录当前的最大因子序列和值

#include <vector>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int vmax=-1,cmax=0;
vector<int> cv;
vector<vector<int>> ans;
bool comp(const vector<int> a,const vector<int> b)
{
    int i=0,len=(a.size()<b.size())?a.size():b.size();
    while(i<len&&a[i]==b[i]) ++i;
    return a[i]>b[i];
}
void Calculate(int n,int k, int p,int s)
{
    if(n<0||k<0)return;
    if(n==0&&k==0)
    {
        if(vmax<cmax)
        {
            ans.clear();
            vmax=cmax;
            ans.push_back(cv);
        }
        else if(vmax==cmax)
            ans.push_back(cv);
        return;
    }
    for(int i=s;i>=1;--i)
    {
        cv.push_back(i);
        cmax+=i;
        Calculate(n-(int)pow(i*1.0,p*1.0),k-1,p,i);
        cv.pop_back();
        cmax-=i;
    }
}
int main()
{
    int n,k,p;
    cin>>n>>k>>p;
    Calculate(n,k,p,n);
    if(!ans.size()){cout<<"Impossible";return 0;}
    sort(ans.begin(),ans.end(),comp);
    cout<<n<<" = ";
    for(auto it=ans[0].begin();it!=ans[0].end();++it)
        it!=ans[0].end()-1?cout<<*it<<"^"<<p<<" + ":cout<<*it<<"^"<<p;
}

以下2017.9.7更新

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
int maxsum=-1;
vector<int> ans;
void solve(vector<int> &ans,vector<int> &rp,int sum,int cur,int n,int k,int p)
{
    if(n<0||k<0) return;
    if(k==0&&n==0&&(sum>maxsum||(sum==maxsum&&ans<rp)))
    {
        ans=rp;
        maxsum=sum;
        return;
    }
    for(int i=cur;i>=1;--i)
    {
        rp.push_back(i);
        solve(ans,rp,sum+i,i,n-pow(i*1.0,p*1.0),k-1,p);
        rp.pop_back();
    }
}
int main()
{
    int n,k,p;
    cin>>n>>k>>p;
    vector<int> rp;
    solve(ans,rp,0,n,n,k,p);
    if(ans.size())
    {
        printf("%d = ",n);
        for(auto it=ans.begin();it!=ans.end();++it)
        {
            printf("%d^%d",*it,p);
            if(it!=ans.end()-1) printf(" + ");
        }
    }
    else
        printf("Impossible");
    return 0;
}