Palindrome Numbers
Time Limit: 2 Seconds Memory Limit: 65536 KB
A palindrome is a word, number, or phrase that reads the same forwards as backwards. For example, the name "anna" is a palindrome. Numbers can also be palindromes (e.g. 151 or 753357).
Additionally numbers can of course be ordered in size. The first few palindrome
numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, ...
The number 10 is not a palindrome (even though you could write it as 010) but a zero as leading digit is not allowed.
Input
The input consists of a series of lines with each line containing one integer value i (1<= i <= 2*10^9 ). This integer value i indicates the index of the palindrome number that is to be written to the output, where index 1 stands for the first palindrome number
(1), index 2 stands for the second palindrome number (2) and so on. The input is terminated by a line containing 0.
Output
For each line of input (except the last one) exactly one line of output containing a single (decimal) integer value is to be produced. For each input value i the i-th palindrome number is to be written to the output.
Sample Input
1
12
24
0
Sample Output
1
33
151
题意：求第i个回文数字；
解析：
位数为1的回文个数：9
位数为2的回文个数：9
位数为3的回文个数：90
位数为4的回文个数：90
位数为5的回文个数：900
位数为6的回文个数：900
位数为7的回文个数：9000
位数为8的回文个数：90000
.……
位数为n的回文个数：9*（10^((n-1)/2)）,
将这些回文数拆分一半，可以发现
---------------------------------------------------------1位
1 2
3 4 5
6 7
8 9
---------------------------------------------------------2位
1 2
3 4 5
6 7
8 9
---------------------------------------------------------3位
10 11
12 13 14
15 16
17 18 19
20 21
22 23 24
25 26
27 28 29
……
90 91
92 93 94
95 96
97 98 99
----------------------------------------------------------4位
10 11
12 13 14
15 16
17 18 19
20 21
22 23 24
25 26
27 28 29
……
90 91
92 93 94
95 96
97 98 99
---------------------------------------------------------5位
100 101
102 103
104 105
106 107
108 109
110 111
112 113
114 115
116 117
118 119
……
990 991
992 993
994 995
996 997
998 999
---------------------------------------------------------6位
……
参考代码：
#include <iostream>
using namespace std;
typedef long long ll;
ll pow(int a,int b){
ll ans = 1;
for (int i=0;i<b;i++){
ans*=a;
}
return ans;
}
int main(){
ll n;
while (cin>>n&&n){
int count=1;
int q=9;
while (n>q){
n-=q;
count++;
if (n>q){
n-=q;
count++;
}
else
break;
q*=10;
}
//cout<<"n="<<n<<endl;
int k=(count-1)/2;
//cout<<"k="<<k<<endl;
ll ans=pow(10,k)+n-1;
//cout<<"ans="<<ans<<endl;
//cout<<"count="<<count<<endl;
int a[20],s=0;
cout<<ans;
while (ans){
a[s++]=ans%10;
ans/=10;
}
for (int j=count/2-1;j>=0;j--){
if (j<0)
break;
cout<<a[s-1-j];
}
cout<<endl;
}
return 0;
}
参考代码：