Additive number

最新推荐文章于 2020-03-08 12:11:57 发布

jeason29

最新推荐文章于 2020-03-08 12:11:57 发布

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本文链接：https://blog.youkuaiyun.com/jeason29/article/details/49912681

本文介绍了一种用于识别加性数的算法实现。加性数是指一个正整数，其数字能够形成至少包含三个数的加性序列。文章详细解释了加性数的概念，并提供了一个C++实现的示例代码，该代码能够判断一个给定的字符串表示的整数是否为加性数。

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Additive number is a positive integer whose digits can form additive sequence.

A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.

For example:
"112358" is an additive number because the digits can form an additive sequence: 1, 1, 2, 3, 5, 8.

1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8

"199100199" is also an additive number, the additive sequence is: 1, 99, 100, 199.

1 + 99 = 100, 99 + 100 = 199

Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

Given a string represents an integer, write a function to determine if it's an additive number.

#include <vector>
#include <iostream>
#include <string>
using namespace std;

class Solution{
public:
        vector<int> additiveNum(int start, int end){
                vector<int> ret;                                                        
                for(int i = start; i <= end; i++)
                        if(isAdditive(i))
                                ret.push_back(i);
                return ret;
        } 
        bool isAdditive(int num){
                string s = to_string(num);
                if(s.length() < 3) return false;
                for(int i = 2; i/2 <= s.length()-i; i++){        // i== length of sum of two first string
                      for(int j = 1; j <= i-1; j++){                        // j== length of first string
                                string s1 = s.substr(0,j);                        // s1: 0,j-1; s2: j,i-1
                                string s2 = s.substr(j,i-j);
                                if(s1[0]=='0' || s2[0]=='0') break;        // 0XX is not allowed
                                int i1 = stoi(s1);
                                int i2 = stoi(s2);
                                int i3 = i1+i2;
                                string s3 = to_string(i3);
                                if(s.find(s3) != i) break;
                                if(helper(s2,s3,s.substr(j))) return true;. more info on 1point3acres.com
                        }
                }
                return false;
        }

        bool helper(string s1, string s2, string s){
                if(s.compare(s1+s2)==0) return true;
                int i1 = stoi(s1); int i2 = stoi(s2);
                int i3 = i1+i2;
                string s3 = to_string(i3);
                int pos = s1.length()+s2.length();
                if(s.find(s3) != pos) return false;
                else return helper(s2,s3,s.substr(s1.length()));. 1point3acres.com/bbs
        }
};

int main(){
        Solution soln;
        vector<int> a = soln.additiveNum(10000,1000000);
        for(int i = 0; i < a.size(); i++). 
                cout<<a[i]<<endl;
}