PAT_甲級_1136 A Delayed Palindrome (20point(s)) (C++)【簽到題/字符串處理】

目錄

1，題目描述

 題目大意

2，思路

3，AC代碼

4，解題過程

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1，題目描述

• palindrome：迴文數；

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.

 題目大意

判斷一個數在10次操作（將這個數字與逆序的數字相加，並將結果作爲新的數字）之內能否變爲迴文數。

 

2，思路

1，bool isPalindrome(string s)：判斷數字（字符串形式）是否爲迴文數；

2，string add(string a)：迭代操作，a與a的逆序相加；

 

3，AC代碼

#include<bits/stdc++.h>
using namespace std;
bool isPalindrome(string s){
    int i = 0, j = s.size()-1;
    while(i < j && s[i] == s[j]){
        i++;j--;
    }
    if(i < j) return false;
    else return true;
}
string add(string a){
    string b = a, c = "";       //c = a + b
    reverse(b.begin(), b.end());
    int carry = 0;
    for(int i = a.size()-1; i >= 0; i--){
        int tem = (a[i]-'0') + (b[i]-'0') + carry;
        if(tem >= 10){          // !!!>=
            tem -= 10;
            carry = 1;
        }else carry = 0;
        c += ('0' + tem);
    }
    if(carry == 1) c += '1';
    reverse(c.begin(), c.end());
    cout<<a<<" + "<<b<<" = "<<c<<endl;
    return c;
}
int main(){
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif // ONLINE_JUDGE
    string s;
    cin>>s;
    int step = 10;
    while(step--){
        if(isPalindrome(s)){
            cout<<s;
            printf(" is a palindromic number.");
            return 0;
        }
        s = add(s);
    }
    printf("Not found in 10 iterations.");
    return 0;
}

4，解題過程

一發入魂o(*￣▽￣*)ブ