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题目链接:https://leetcode.com/problems/additive-number/
题目描述:一串数字是Additive number,必须存在一个划分将这一串数字分成n个数字,n必须大于3,并且除了前两个数字之外,每个数字都是前两个数字之和,此外,不存在包含前导零的数字。
题目分析:深度搜索
代码实现:
#include <map> #include <vector> #include <algorithm> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <iostream> #include <stack> using namespace std; bool judge(string a, string b, string sum) { //cout<<a<<", "<<b<<", "<<sum<<endl; int lena = a.length(); int lenb = b.length(); int lens = sum.length(); if (lens < lena || lens < lenb) return 0; int c = 0; while(lena > 0 && lenb > 0) { int tmp = a[lena-1] + b[lenb-1] - ‘0‘ - ‘0‘ + c; c = tmp / 10; tmp = tmp % 10; if (sum[lens-1] != tmp+‘0‘) return 0; lena--; lenb--; lens--; } while(lena > 0) { int tmp = a[lena-1] + c -‘0‘; c = tmp / 10; tmp = tmp % 10; if (sum[lens-1] != tmp+‘0‘) return 0; lena--; lens--; } while(lenb > 0) { int tmp = b[lenb-1] + c - ‘0‘; c = tmp /10; tmp = tmp %10; if (sum[lens-1] != tmp+‘0‘) return 0; lenb--; lens--; } if ((c==0 && lens==0) || (lens==1 && sum[0]==c+‘0‘)) return 1; return 0; } bool df(string a, string b, string num) { int lena = a.length(); int lenb = b.length(); int lenn = num.length(); int len = max(lena, lenb); if (!lenn) return 1; if (len > lenn) return 0; if ((a[0] == ‘0‘ && lena>1) || (b[0] == ‘0‘ && lenb>1) || (num[0] == ‘0‘ && lenn>1)) return 0; bool ret = judge(a, b, num.substr(0, len)); //cout<<a<<", "<<b<<", "<<num.substr(0, len)<<", ret:"<<ret<<endl; if (ret) { if (df(b, num.substr(0, len), num.substr(len))) return 1; } if (len+1 > lenn) return 0; ret = judge(a, b, num.substr(0, len+1)); //cout<<a<<", "<<b<<", "<<num.substr(0, len+1)<<", ret:"<<ret<<endl; if (ret) { if (df(b, num.substr(0, len+1), num.substr(len+1))) return 1; } return 0; } bool isAdditiveNumber(string num) { int len = num.length(); for (int i = 0; i < len-2; i++) { for (int j = i+1; j < len-1; j++) { if (df(num.substr(0, i+1), num.substr(i+1, j-i), num.substr(j+1))) return 1; } } return 0; } int main() { string s; while(cin>>s) { cout<<s<<" is "<<isAdditiveNumber(s)<<endl; } return 0; }
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原文地址:http://www.cnblogs.com/li-xingtao/p/4981254.html