ZOJ-1334 进制转换

1334：给出源数 给出源数的基 和目标数的基 求目标数
目标数最长7位 右对齐显示

Sample Input


1111000 2 10
  1111000 2 16
2102101  3 10
2102101 3   15
12312 4   2
1A   15 2
   1234567 10 16
  ABCD 16 15

Sample Output

    120
     78
   1765
    7CA
  ERROR
  11001
12D687
   D071


思路:将源数转换成10进制，再转成目标进制
char res[8] 数组存字符串一定要记得给末尾\0留一位！！！

#include<iostream>
#include<string>
#include<math.h>
#include<stdio.h>
#include<memory.h>
using namespace std;

char ref[16]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
int count=0;
char res[8];

void ten2n(int num,int base)
{
	if(num/base!=0)
		ten2n(num/base,base);
	
	if(count<7)
		res[count++]=ref[num%base];
	else if(count++==7)
		strcpy(res,"ERROR");

}

int n2ten(char * num,int base)
{
	int sum=0;
	int val;
	int len = strlen(num);
	for(int i=0;i<len;i++)
	{
		if(num[i]>='0'&&num[i]<='9')
			val=num[i]-'0';
		else if(num[i]>='A'&&num[i]<='F')
			val=num[i]-'A'+10;
		sum+=val*pow((double)base,len-1-i);
	}
	return sum;
}

int main()
{
	char src[100];
	int ibase;
	int obase;
	while(cin>>src)
	{
		cin>>ibase;
		cin>>obase;
		count=0;
		memset(res,0,sizeof(char)*8);
		ten2n(n2ten(src,ibase),obase);
		memset(src,0,sizeof(char)*100);
		printf("%7s\n",res);
	}


}