The reverse and add function starts with a number, reverses its digits and adds the reverse to the original. If the sum is not a palindrome (meaning it does not give the same number read from left to right and right to left), we repeat this procedure until it does.
For example, if we start with 195 as the initial number, we get 9,339 as the resulting palindrome after the fourth addition:
| 195 |
|
786 |
|
1,473 |
|
5,214 |
| 591 |
|
687 |
|
3,741 |
|
4,125 |
| + --- |
|
+ --- |
|
+ --- |
|
+ --- |
| 786 |
|
1,473 |
|
5,214 |
|
9,339 |
This method leads to palindromes in a few steps for almost all of the integers. But there are interesting exceptions. 196 is the first number for which no palindrome has been found. It has never been proven, however, that no such palindrome exists.
You must write a program that takes a given number and gives the resulting palindrome (if one exists) and the number of iterations/additions it took to find it.
You may assume that all the numbers used as test data will terminate in an answer with less than 1,000 iterations (additions), and yield a palindrome that is not greater than 4,294,967,295.
The first line will contain an integer
N ( 0 <
N
100), giving the number of test cases, while the next
N lines each contain a single integer
P whose palindrome you are to compute.
For each of the
N integers, print a line giving the minimum number of iterations to find the palindrome, a single space, and then the resulting palindrome itself.
3
195
265
750
4 9339
5 45254
3 6666
一個簡單的處理問題 直接用加法模板 過的 這個服務器由于是gcc標準 所有很多以前用習慣的函數 都不支持 如 strrev 還得自己寫個
不過還好 不難
貼上代碼
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
const int maxn = 110;
void add(char a[],char b[],char back[])
{
int i,j,k,up,x,y,z,l;
char *c;
if (strlen(a)>strlen(b)) l=strlen(a)+2; else l=strlen(b)+2;
c=(char *) malloc(l*sizeof(char));
i=strlen(a)-1;
j=strlen(b)-1;
k=0;up=0;
while(i>=0||j>=0)
{
if(i<0) x='0'; else x=a[i];
if(j<0) y='0'; else y=b[j];
z=x-'0'+y-'0';
if(up) z+=1;
if(z>9) {up=1;z%=10;} else up=0;
c[k++]=z+'0';
i--;j--;
}
if(up) c[k++]='1';
i=0;
c[k]='\0';
for(k-=1;k>=0;k--)
back[i++]=c[k];
back[i]='\0';
}
char *strrev(char *s)
{
if (s == NULL || s[0] == '\0')
return s;
for (char t, *p = s, *q = s + strlen(s) - 1; p < q; p++, q--)
t = *p, *p = *q, *q = t;
return s;
}
bool isReverse(char str[])
{
char *temp;
temp = (char*)malloc(sizeof(char)*(strlen(str) +3));
strcpy(temp,str);
strrev(temp);
//temp = strrev(str);
if(strcmp(str,temp)==0)
return true;
return false;
}
int main()
{
char str1[maxn],str2[maxn],back[maxn];
int i,n,num,T;
scanf("%d\n",&T);
while(T--)
{
i = 0;
gets(str1);
strcpy(str2,str1);
strrev(str2);
strcpy(back,"0");
while(!isReverse(str1))
{
add(str1,str2,back);
strcpy(str1,back);
strcpy(str2,str1);
strrev(str2);
i ++;
}
printf("%d %s\n",i,str1);
}
return 0;
}
posted on 2010-07-23 19:43
付翔 閱讀(307)
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ACM 數據結構