（算法练习）——数制转换

要求：
http://codeup.cn/problem.php?cid=100000579&pid=1
说明：
根据提示，用字符串来保存每次输入的数据（因为超过10进制可能有字母）；在输出的时候对应挨个输出，遇到大于等于10就手动转成字母= =
另外感觉pow挺好用的
代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
struct record{
	int num;
	int ansrecord[40];
}recorda[1000];

int main(){
	char n[100];
	int a,b; 
	int signal = 0;
	while(scanf("%d %s %d",&a,n,&b) != EOF){
		int len = strlen(n);
		long long sum = 0;
		for(int i = len - 1;i >= 0;i--){
			if(n[i] == 'A' || n[i] == 'a' ){
				sum = sum + (10)* pow(a,len - i -1);
			}
			else if(n[i] == 'B' || n[i] == 'b'){
				sum = sum + (11)* pow(a,len - i -1);
			}
			else if(n[i] == 'C' || n[i] == 'c'){
				sum = sum + (12)* pow(a,len - i -1);
			}
			else if(n[i] == 'D' || n[i] == 'd'){
				sum = sum + (13)* pow(a,len - i -1);
			}
			else if(n[i] == 'E' || n[i] == 'e'){
				sum = sum + (14)* pow(a,len - i -1);
			}
			else if(n[i] == 'F' || n[i] == 'f'){
				sum = sum + (15)* pow(a,len - i -1);
			}
			else{
				sum = sum + (n[i] - '0')* pow(a,len - i -1);
			}	
		}
		int numa = 0;
		do{
			recorda[signal].ansrecord[numa++] = sum %b;
			sum /= b;
		}while(sum != 0);
		recorda[signal].num = numa;
		signal++;
		
	}
	for(int i = 0;i <signal;i++){
		for(int j = recorda[i].num - 1;j >= 0;j--){
			
			if(recorda[i].ansrecord[j] == 10){
				printf("A");
			}
			else if(recorda[i].ansrecord[j] == 11){
				printf("B");
			}
			else if(recorda[i].ansrecord[j] == 12){
				printf("C");
			}
			else if(recorda[i].ansrecord[j] == 13){
				printf("D");
			}
			else if(recorda[i].ansrecord[j] == 14){
				printf("E");
			}
			else if(recorda[i].ansrecord[j] == 15){
				printf("F");
		    }
		    else{
		    	printf("%d",recorda[i].ansrecord[j]);
			}	
		}
		printf("\n");
	}
}