力扣 2698. 求一个整数的惩罚数

题目来源：https://leetcode.cn/problems/find-the-punishment-number-of-an-integer/

 C++题解：单独写了个函数判断是否可以分割，判断方法为暴力法。然后更新惩罚数。

bool fin(vector m, int tar) {
    int len = m.size();
    int sum = 0;
    vector tmp(0);
    for(int ii = 0; ii < len; ii++) {
        sum = sum + m[ii];
    }
    tmp.push_back(sum);
    if(len == 4){
        sum = m[0]*10 + m[1] + m[2] + m[3]; tmp.push_back(sum);
        sum = m[0]*10 + m[1] + m[2]*10 + m[3]; tmp.push_back(sum);
        sum = m[0] + m[1]*10 + m[2] + m[3]; tmp.push_back(sum);
        sum = m[0] + m[1] + m[2]*10 + m[3]; tmp.push_back(sum);
        sum = m[0]*100 + m[1]*10 + m[2] + m[3]; tmp.push_back(sum);
        sum = m[0] + m[1]*100 + m[2]*10 + m[3]; tmp.push_back(sum);
    }
    if(len == 5){
        sum = m[0]*10 + m[1] + m[2] + m[3] + m[4]; tmp.push_back(sum);
        sum = m[0]*10 + m[1] + m[2]*10 + m[3] + m[4]; tmp.push_back(sum);
        sum = m[0]*10 + m[1] + m[2] + m[3]*10 + m[4]; tmp.push_back(sum);
        sum = m[0]*10 + m[1] + m[2]*100 + m[3]*10 + m[4]; tmp.push_back(sum);
        sum = m[0] + m[1]*10 + m[2] + m[3] + m[4]; tmp.push_back(sum);
        sum = m[0] + m[1]*10 + m[2] + m[3]*10 + m[4]; tmp.push_back(sum);
        sum = m[0] + m[1] + m[2]*10 + m[3] + m[4]; tmp.push_back(sum);
        sum = m[0] + m[1] + m[2] + m[3]*10 + m[4]; tmp.push_back(sum);
        sum = m[0]*100 + m[1]*10 + m[2] + m[3] + m[4]; tmp.push_back(sum);
        sum = m[0]*100 + m[1]*10 + m[2] + m[3]*10 + m[4]; tmp.push_back(sum);
        sum = m[0] + m[1]*100 + m[2]*10 + m[3] + m[4]; tmp.push_back(sum);
        sum = m[0] + m[1] + m[2]*100 + m[3]*10 + m[4]; tmp.push_back(sum);
        sum = m[0]*1000 + m[1]*100 + m[2]*10 + m[3] + m[4]; tmp.push_back(sum);
        sum = m[0] + m[1]*1000 + m[2]*100 + m[3]*10 + m[4]; tmp.push_back(sum);        
    }
    if(len == 6){
        sum = m[0]*10 + m[1] + m[2] + m[3] + m[4] + m[5]; tmp.push_back(sum);
        sum = m[0] + m[1]*10 + m[2] + m[3] + m[4] + m[5]; tmp.push_back(sum);
        sum = m[0] + m[1] + m[2]*10 + m[3] + m[4] + m[5]; tmp.push_back(sum);
        sum = m[0] + m[1] + m[2] + m[3]*10 + m[4] + m[5]; tmp.push_back(sum);
        sum = m[0] + m[1] + m[2] + m[3] + m[4]*10 + m[5]; tmp.push_back(sum);
        sum = m[0]*100 + m[1]*10 + m[2] + m[3] + m[4] + m[5]; tmp.push_back(sum);
        sum = m[0]*100 + m[1]*10 + m[2] + m[3]*10 + m[4] + m[5]; tmp.push_back(sum);
        sum = m[0]*100 + m[1]*10 + m[2] + m[3] + m[4]*10 + m[5]; tmp.push_back(sum);
        sum = m[0]*100 + m[1]*10 + m[2] + m[3]*100 + m[4]*10 + m[5]; tmp.push_back(sum);
        sum = m[0] + m[1]*100 + m[2]*10 + m[3] + m[4] + m[5]; tmp.push_back(sum);
        sum = m[0] + m[1]*100 + m[2]*10 + m[3] + m[4]*10 + m[5]; tmp.push_back(sum);
        sum = m[0] + m[1] + m[2]*100 + m[3]*10 + m[4] + m[5]; tmp.push_back(sum);
        sum = m[0]*10 + m[1] + m[2]*100 + m[3]*10 + m[4] + m[5]; tmp.push_back(sum);
        sum = m[0] + m[1] + m[2] + m[3]*100 + m[4]*10 + m[5]; tmp.push_back(sum);
        sum = m[0]*10 + m[1] + m[2] + m[3]*100 + m[4]*10 + m[5]; tmp.push_back(sum);
        sum = m[0] + m[1]*10 + m[2] + m[3]*100 + m[4]*10 + m[5]; tmp.push_back(sum);
        sum = m[0]*1000 + m[1]*100 + m[2]*10 + m[3] + m[4] + m[5]; tmp.push_back(sum);
        sum = m[0]*1000 + m[1]*100 + m[2]*10 + m[3] + m[4]*10 + m[5]; tmp.push_back(sum);
        sum = m[0] + m[1]*1000 + m[2]*100 + m[3]*10 + m[4] + m[5]; tmp.push_back(sum);   
        sum = m[0] + m[1] + m[2]*1000 + m[3]*100 + m[4]*10 + m[5]; tmp.push_back(sum);  
        sum = m[0]*10 + m[1] + m[2]*1000 + m[3]*100 + m[4]*10 + m[5]; tmp.push_back(sum);     
        sum = m[0]*10000 + m[1]*1000 + m[2]*100 + m[3]*10 + m[4] + m[5]; tmp.push_back(sum);  
        sum = m[0] + m[1]*10000 + m[2]*1000 + m[3]*100 + m[4]*10 + m[5]; tmp.push_back(sum);      
    }
    //cout< m(0);
            while(tem > 0){                
                m.push_back(tem % 10);
                tem = (tem - tem % 10) / 10;
            }
            reverse(m.begin(), m.end());
            if(fin(m, ii)) res = res + ii * ii;            
        }
        return res;
    }
};
//45 3503  297-184768 414-772866

C++题解2：使用DFS深度优先搜索算法，还没学，先放着。

class Solution {
public:
    string s;
    int punishmentNumber(int n) {
        int ans = 0;

        function dfs = [&](int x, int start, int len, int sum) -> bool {
            /* 提前返回 */
            if (sum > x) {
                return false;
            }
            if (start == len) {
                if (sum == x) { /* 处理到最后一个, 可以等于原数 */
                    return true;
                }
                return false;
            }
            int tmp = 0;
            for (int j = start; j < len; j++) { /* 枚举分割点 */
                tmp = tmp * 10 + s[j] - '0';
                if (dfs(x, j + 1, len, sum + tmp)) {
                    return true;
                }
            }
            return false;
        };

        for (int i = 1; i <= n; i++) {
            s = to_string(i * i);
            if (dfs(i, 0, s.size(), 0)) {
                ans += i * i;
            }
        }
        return ans;
    }
};

作者：linge32
链接：https://leetcode.cn/problems/find-the-punishment-number-of-an-integer/solutions/2278318/c-dfs-by-liu-xiang-3-bhzl/
来源：力扣（LeetCode）
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