Inspired by IBM’s Blue Gene project, the CEO of Universal Biological Machinery (UBM), has called on you, UBM’s top software engineer, to develop a program that will calculate the mutation of the Areopagus-virus, a virus discovered on Mars by your company’s privately-subsidized (top-secret) space program.
Input
Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets.
A single data set has 3 components:

Start line - A single line, “START N”, where 1 <= N <= 20.
Viral code - A sequence of N alphanumeric characters. Alphanumeric characters will consist of an uppercase letter (A-Z) or a digit (0-9).
End line - A single line, “END”

Following the final data set will be a single line, “ENDOFINPUT”.
Output
For each data set, there will be exactly one output set, and there will be no blank lines separating output sets.

A single output set consists of a single line of the viral code after it has stabilized (through mutating).

The viral code will mutate according to the following rules:

Initially the first viral segment to mutate begins with the first alphanumeric character of the viral code and ends with the rightmost alphanumeric character of the code.
If the first alphanumeric character of a viral segment is a letter (A-Z), that alphanumeric character is considered “unstable”, and will mutate into n, where n is the number of mutations that occur to the viral segment immediately to the unstable alphanumeric character’s right (see #5), unless n is greater than 9, in which case the unstable alphanumeric character will mutate into n % 10. Also, if there is no viral segment immediately to the right of the unstable alphanumeric character, the unstable alphanumeric character will mutate into 0.
If the first alphanumeric character of a viral segment, n, is a positive number (1-9), that alphanumeric character is also considered “unstable”, and will mutate into n-1. It also causes the viral segment beginning with the alphanumeric character n alphanumeric characters to its right and ending with the rightmost alphanumeric character of the viral code to mutate. If there is no alphanumeric character n alphanumeric characters to its right, then the viral segment immediately to its right (see #5), if one exists, will mutate.
If the first alphanumeric character of a viral segment is 0, that alphanumeric character is considered “stable”, and will not mutate (the alphanumeric character will remain a 0 and a mutation will not be considered to have occurred).
A viral segment immediately to the right of an alphanumeric character begins with the alphanumeric character one position to its right and ending with the rightmost alphanumeric character of the viral code.

Sample Input
START 1
A
END
START 4
A1B2
END
START 15
A3B2CCC4AD1232R
END
START 15
0ABCDEFGHIJKLMN
END
START 11
ABCDEFGHIJK
END
START 10
9AAAAAAAAA
END
ENDOFINPUT

Sample Output
0
3011
82B26543AD11310
0ABCDEFGHIJKLMN
09876543210
8AAAAAAAA0
题意分析：
就是给你一串字符串，如果字符串是字母开头的，此字母就变成后面变异数的总和。
如果字符串的开头是数字n，此数字变成n-1，然后从后面第n个开始变异。
如果字符串以0开头，就不发生变异，直接输出。
解题思路：
就是一个递归的过程
1.字符串是字母开头时,直接向后移一位，进行递归。
2.开头是数字n时，就变成n-1,如果后面还有n位，就移到n位再继续转变，否则右移一位开始转变，进行递归。
3.当开头为0时，直接结束递归。

#include <stdio.h>
#include <string.h>
s
int digui(int t); 
int n,t=0;
char a[25];
char str[25];
int main()
{
    
    
    
    while (scanf("%s",str)) 
	{
    
    
        if (strcmp(str,"ENDOFINPUT")==0)
        {
    
    
        	break;
		}
        scanf("%d%s%s",&n,a,str);
        digui(0);
        printf("%s\n",a);
    }
    return 0;
}
int digui(int t)
{
    
    
	int  m; 
    if (a[t]=='0'||t == n) //递归出口 
	{
    
    
        return 0;
    }
    if (a[t]>='A'&&a[t]<='Z') //字母处理 
	{
    
    
        m=digui(t+1);
    	a[t]=(m%10)+'0';
        return  m+1;
    }
    if (a[t]>='1'&&a[t]<='9') //数字处理 
	{
    
    
		a[t]--;
        if (t+a[t]-'0'+1<n) 
		{
    
    
            m=digui(t+a[t]-'0'+1);
            return m+1;
        }
        else 
		{
    
    
            m=digui(t+1);
            return m+1;
        }
    }
}