PAT 1136 A Delayed Palindrome[简单]

1136 A Delayed Palindrome （20 分）

Consider a positive integer

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:

Sample Output 1:

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

Sample Input 2:

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.

题目大意：给出一个数，是否回文？否则将其反转然后和原数相加，直到加到第10次未出现或者出现。

#include <iostream>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstdlib>
using namespace std;

//但是属于哪一条地铁，怎么去表示呢？
bool pd(string s){
    int sz=s.size();
    for(int i=0;i<sz/2;i++){
        if(s[i]!=s[sz-i-1])
            return false;
    }
    return true;
}
int toNum(string s){
    int a=0;
    for(int i=0;i<s.size();i++){
        a=a*10+(s[i]-'0');
    }
    return a;
}
string toStr(int a){
    string s;
    while(a!=0){
        s+=(a%10+'0');
        a/=10;
    }
    reverse(s.begin(),s.end());
    return s;
}
int main(){
    string s;
    cin>>s;
    string t=s;
    //reverse(s[0],s[0]+s.size());//那个反转函数是啥来着???
    //s.reverse();//这个调用不正确
    reverse(s.begin(),s.end());//t是第一个数，s是第二个数.
    bool flag=false;
    int a,b,c;
    for(int i=0;i<10;i++){
        if(pd(s)){//如果这个数本来就是答案。
            flag=true;break;
        }
        a=toNum(t);
        b=toNum(s);
        c=a+b;
        cout<<t<<" + "<<s<<" = "<<c<<'
';
        s=toStr(c);
        t=s;
        reverse(s.begin(),s.end());
    }
    if(flag){
        cout<<t<<" is a palindromic number.";
    }else{
        cout<<"Not found in 10 iterations.";
    }
    return 0;
}

//本来是这么写的，但是提交只有18分，最后一个测试点过不去：答案错误。

//我明白了，这个的考点是大数相加(1K位)，而我却将其转换成int，实在是太不好了。

#include <iostream>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstdlib>
using namespace std;

string add(string a){
    string b=a;
    reverse(b.begin(),b.end());
    int jin=0,s;
    for(int i=a.size()-1;i>=0;i--){
        s=(a[i]-'0')+(b[i]-'0')+jin;
        if(s>=10){
            s%=10;
            jin=1;
        }else jin=0;
        a[i]=s+'0';
    }
    if(jin==1){
        a="1"+a;
    }
    return a;
}

int main(){
    string s;
    cin>>s;
    string t=s;
    //reverse(s[0],s[0]+s.size());//那个反转函数是啥来着???
    //s.reverse();//这个调用不正确
    reverse(s.begin(),s.end());//t是第一个数，s是第二个数.
    bool flag=false;
    for(int i=0;i<10;i++){
        if(t==s){//如果这个数本来就是答案。
            flag=true;break;
        }
        cout<<t<<" + "<<s<<" = "<<add(t)<<'
';
        t=add(t);
        s=t;
        reverse(s.begin(),s.end());
    }
    if(flag){
        cout<<t<<" is a palindromic number.";
    }else{
        cout<<"Not found in 10 iterations.";
    }
    return 0;
}

//终于正确了，这水题花了我好长时间，哭唧唧！你看清题好吗？1000位的数字就是模拟大数相加啊！！！