1136 A Delayed Palindrome（41行代码+超详细注释）

分数 20

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作者 CHEN, Yue

单位 浙江大学

Consider a positive integer N written in standard notation with k+1 digits ai​ as ak​⋯a1​a0​ with 0≤ai​<10 for all i and ak​>0. Then N is palindromic if and only if ai​=ak−i​ for all i. Zero is written 0 and is also palindromic by definition.

Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. Such number is called a delayed palindrome. (Quoted from https://en.wikipedia.org/wiki/Palindromic_number )

Given any positive integer, you are supposed to find its paired palindromic number.

Input Specification:

Each input file contains one test case which gives a positive integer no more than 1000 digits.

Output Specification:

For each test case, print line by line the process of finding the palindromic number. The format of each line is the following:

A + B = C

where A is the original number, B is the reversed A, and C is their sum. A starts being the input number, and this process ends until C becomes a palindromic number -- in this case we print in the last line C is a palindromic number.; or if a palindromic number cannot be found in 10 iterations, print Not found in 10 iterations. instead.

Sample Input 1:

97152

Sample Output 1:

97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.

Sample Input 2:

196

Sample Output 2:

196 + 691 = 887
887 + 788 = 1675
1675 + 5761 = 7436
7436 + 6347 = 13783
13783 + 38731 = 52514
52514 + 41525 = 94039
94039 + 93049 = 187088
187088 + 880781 = 1067869
1067869 + 9687601 = 10755470
10755470 + 07455701 = 18211171
Not found in 10 iterations.

代码长度限制

16 KB

时间限制

400 ms

内存限制

64 MB

#include
using namespace std;
string add(string a,string b){//将a，b换成十进制相加 
    int cc=0,len=a.size();//cc表示进位一开始为0 
    string t;
    for(int i=len-1;i>=0;i--){//从个位开始相加 
        int x=(a[i]-'0'+b[i]-'0'+cc)%10;//相加取低位 
        cc=(a[i]-'0'+b[i]-'0'+cc)/10;//每次的进位 
        t=to_string(x)+t;//将每次的低位以字符串形式保存在t里 
    }
    if(cc==1)t="1"+t;//最后一次的进位为1则加上 
    return t;
}
bool ispal(string s){//判断是否为回文数 
    int len=s.size();
    for(int i=0;i         if(s[i]!=s[len-1-i])return false;
    }
    return true;
}
int main(){
    string s;
    cin>>s;
    if(ispal(s)){//若输入的数就是回文数则直接输出 
        printf("%s is a palindromic number.\n",s.c_str());
        return 0;
    }
    for(int i=1;i<=10;i++){//否则 
        string ss=s;
        reverse(ss.begin(),ss.end());//得到s的反转 
        string res=add(s,ss);//将s和ss相加 
        printf("%s + %s = %s\n",s.c_str(),ss.c_str(),res.c_str());//输出 
        if(ispal(res)){//若是回文数则输出并中断 
            printf("%s is a palindromic number.\n",res.c_str());
            break;
        }
        if(i==10)cout<<"Not found in 10 iterations."<         s=res;//更新s 
    }
    return 0;
}