Codeforces Gym 101164I Cubes

Problem I

Cubes

Input File: I.in
Output File: standard output
Time Limit: 0.5 seconds (C/C++)
Memory Limit: 256 megabytes
Write a program that takes a natural number N and decomposes it as a sum of the minimum number of
exact natural cubes. The program should find m1,m2, …,mk, such that eachmi is a natural number,
m1
3
+ m23+… + mk3= N, andk is minimal.
Input
The only line of the input file contains the numberN (1 ≤N ≤ 44,777,444).
Output
Your program should write exactly two lines. The first line contains the numberk - the minimum
number of natural cubes. The second line contains kspace-separated natural numbers - that
raised to the power of 3 sum to N.

Sample inputSample output42 7
2 2 2 2 2 1 143 3
3 2 2

题目大意：给定一个数n，要求将n分成k个数的立方和，使得k最小。

思路：该题有多种做法，可以采用BFS的方法，也可以采用DFS的方法。

BFS：

#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<iostream>#include<cmath>#include<cstdlib>#include<queue>using namespace std;const int MAXN=50000010;int dis[MAXN];int ans[1000];queue <int> q;bool p;int work(int n,int required,int MAX){if (dis[n]<1000 && !p){return dis[n];}if (!n){return 0;}if (required<=2 && dis[n]>required){return 3;}int ret=required+1;for (int i=MAX;i>=1;i--){int m=n-i*i*i;if (m<0){continue;}int res=work(m,required-1,i)+1;if (res<ret){ans[required-1]=i;ret=res;}required=ret-1;} return ret;}int main(){int n;scanf("%d",&n);for (int i=0;i<MAXN;i++){dis[i]=1000;}q.push(0);dis[0]=0;while (!q.empty()){int x=q.front();if (dis[x]==3){break;}q.pop();for (int i=1;x+i*i*i<MAXN;i++){int y=x+i*i*i;if (dis[y]>dis[x]+1){dis[y]=dis[x]+1;q.push(y);}}}int num=work(n,100,400);p=true;work(n,num,400);printf("%d\n",num);for (int i=0;i<num;i++){printf("%d ",ans[i]);}    return 0;}

DFS：

#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<iostream>#include<cmath>#include<cstdlib>#include<ctime>using namespace std;const int MAXN=500;int n,tot;int p[MAXN];int num[MAXN];void dfs(int n,int i,int rem){if (n==0){printf("%d\n",tot);for (int i=0;i<tot;i++){printf("%d ",num[i]);}exit(0);}if (rem<=0){return ;}int id=upper_bound(p+1,p+MAXN,n)-(p+1);i=min(i,id);if (n>p[i]*rem){return ;}tot++;while (i>=1){num[tot-1]=i;dfs(n-p[i],i,rem-1);i--;}tot--;}int main(){for (int i=1;i<MAXN;i++){p[i]=i*i*i;}scanf("%d",&n);for (int i=1;;i++){dfs(n,MAXN-1,i);}     return 0;}