Algorithm Training palindrome Number

Algorithm Training palindrome Number
Time limit: 1.0s memory limit: 256.0MB
Submit this Title 1 Jin Sac 2
Problem description
If a number (first not 0) is read from left to right and read from right to left, we call it a palindrome number.
For example: Given a 10 binary number 56, will 56 plus 65 (that is, 56 right-to-left reading), to get 121 is a palindrome number.

Another example: for 10 binary number 87:
step1:87+78 = 165 step2:165+561 = 726
step3:726+627 = 1353 step4:1353+3531 = 4884

One step here is to make an n-ary addition, and the above example uses at least 4 steps to get the palindrome number 4884.

Write a program, given an N (2<=n<=10 or n=16) binary number m (where 16 binary number is 0-9 and a-f), the minimum number of steps can be obtained palindrome numbers.
If the palindrome number cannot be obtained within 30 steps (including 30 steps), the output "impossible!"
Input format
Two lines, N and M
Output format
If you can get palindrome number within 30 steps, Output "step=xx" (without quotation marks), where xx is the number of steps; otherwise output line "impossible!" (without quotation marks)
Sample input
9
87
Sample output
Step=6

http://lx.lanqiao.cn/problem.page?gpid=T74

The law of the addition of the additive adds precision traversal. Flood problem

#include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> using namespace std;
int d[10001];
    int run (int n) {int z=0;
    int l=10000;
    int p[10001];
    memset (P,0,sizeof (p));

    while (d[l]==0) l--;
        for (int i=0;i<=l;i++) {p[i]=d[l-i];
    cout<<d[l-i];


    }//cout<<endl<<endl;
        for (int i=0;i<10000;i++) {d[i]=d[i]+p[i]+z;
        z=d[i]/n;
    D[i]%=n;
    } l=10000;
    while (d[l]==0) l--;
    /* for (int j=l;j>=0;j--) {cout<<d[j];
    } cout<<endl<<endl;*/for (int j=0,i=l;i>=0;i--, J + +) {if (D[i]!=d[j]) return 0;
} return 1;
    } int main () {int n;
        while (cin>>n) {memset (d,0, sizeof (d));
        string S;
        cin>>s;
        int i;
            for (i=0;s[i];i++) {if (s[i]>= ' 0 ' &&s[i]<= ' 9 ') d[i]=s[i]-' 0 '; else d[i]=s[i]-' A ' +10; } for (i=1;i<30;i++) if (run (n)) {cout<< "step=" <<i<<e
                Ndl
            Break } if (i==30) cout<< "impossible!"
    <<endl;
} return 0; }