1103 Integer Factorization (30 分)

最新推荐文章于 2021-09-03 15:25:16 发布

原创最新推荐文章于 2021-09-03 15:25:16 发布 · 109 阅读

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该博客内容涉及一个C++程序，使用递归搜索找到K个整数的组合，使得这些整数的P次幂之和等于给定的目标值N。程序通过平衡栈来存储中间结果，并在过程中优化解决方案。在满足条件的情况下更新最大和以及最优解。测试点2未能通过，可能需要进一步调试代码以找出问题所在。

#include <bits/stdc++.h>
using namespace std;
//FILE *p = fopen("file.txt", "w");
int N, K, P, maxsum = 0, mintop = 20;
stack<int> maxs, s;
stack<int> tmps ,tmpmaxs;
int up;
void dfs(int cnt, int n, int sum, int ans)
{

    if (n != 0)
        s.push(n);

    if (ans > N || n > up || cnt > K || (cnt == K && ans == N && sum < maxsum))
    {
        s.pop();
        return;
    }
    else if (cnt == K && ans == N && sum >= maxsum)
    {
        if (s.top() < mintop)
            mintop = s.top();
        //fprintf(p, "%d %d %d %d\n", cnt, n, sum, ans);
        //printf("%d>=%d\n",sum,maxsum);
        

        if (sum > maxsum||maxs.empty())
            maxs = s;
        else if(sum == maxsum )
        {
            tmps = s;
            tmpmaxs = maxs;
            while (tmps.size() > 0)
            {
                //printf("%d>=%d\n",sum,maxsum);
                if(tmps.top() == tmpmaxs.top())
                {
                    //printf("%d==%d\n",tmps.top(),tmpmaxs.top());
                    tmps.pop();
                    tmpmaxs.pop();
                }
                else if (tmps.top() < tmpmaxs.top()){
                    break;
                }
                    
                else if (tmps.top() > tmpmaxs.top())
                {
                    maxs = s;
                    break;
                }
                
            }
        }
        maxsum = sum;
        s.pop();
        return;
    }
    //fprintf(p, "%d %d %d %d\n", cnt, n, sum, ans);
    if (n == 0)
        for (int i = 1; i <= up; i++)
        {
            if (n + i <= mintop)
                dfs(cnt + 1, n + i, sum + n + i, ans + pow(n + i, P));
            else
                break;
        }
    else
        for (int i = 0; i <= up - n; i++)
        {
            if (n + i <= mintop)
                dfs(cnt + 1, n + i, sum + n + i, ans + pow(n + i, P));
            else
                break;
        }
    if (n != 0)
        s.pop();
    return;
}

int main()
{
    scanf("%d%d%d", &N, &K, &P);

    up = floor(pow(1.0 * N - K + 1, 1.0 / P));

    dfs(0, 0, 0, 0);

    if (maxsum == 0)
        printf("Impossible\n");
    else
    {
        printf("%d = ", N);
        while (maxs.size() > 1)
        {
            printf("%d^%d + ", maxs.top(), P);
            maxs.pop();
        }
        if (maxs.size() == 1)
            printf("%d^%d\n", maxs.top(), P);
    }

    return 0;
}

测试点2过不了