找出这样的正整数：它不是9个不同的正整数的立方和。

#include <iostream>

using namespace std;



#include <stdio.h>

#define MAX_N 100

#define MAX_SQ 20000

int n3Map[MAX_N+1];

int s3Map[MAX_N+1];

int sq3Map[MAX_SQ+1];



unsigned Sqrt3(unsigned n)

{

    unsigned nl=0, nr=MAX_N, nm=(MAX_N>>1);

    for (;nl<nr;nm=(nl+nr+1)>>1)if ((nm*nm*nm)>n)nr=nm-1;else nl=nm;

    return nm;

}



void init()

{

    s3Map[0] = n3Map[0] = sq3Map[0] = 0;

    for (int n=1; n<=MAX_N; ++n)

    {

        n3Map[n] = n*n*n;

        s3Map[n] = s3Map[n-1] + n3Map[n];

    }

    for (int n=1; n<=MAX_SQ; ++n)

    {

        sq3Map[n] = Sqrt3(n);

    }

}



int DFS(int nCnt, int nNums, int nMax=MAX_N)

{

    int n3Root = sq3Map[nCnt];

    if (nMax<=0 || nCnt<=0) return -1;

    if (nNums<=1)

    {

        if (n3Map[n3Root] == nCnt)

            return 1;

    }

    else

    {

        int n=n3Root;

        if (n>nMax) n = nMax;

        for (; s3Map[n]>=nCnt; --n)

        {

            int nRe = DFS(nCnt-n3Map[n], nNums-1, n-1);

            if (nRe == 1)

                return 1;

        }

    }

    return 0;

}



#include <time.h>

int main(int argc, char *argv[])

{

    init();

    int n=27;

    int t = clock();

    int nc=2025-1;

    for (n=2025; n<=MAX_SQ; ++n) //忽略2025前的解

    {

        int nr = DFS(n,9);

        if (nr!=1)

        {

            ++nc;

            printf("%d ",n); //注释掉这句可以获得更快的速度

        }

    }

    t = clock()-t;

    printf("\nTotal= %d \tTime= %d ms\n", nc, t);

    return 0;

}