joj1006（进制转换）

1006: All your base

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	5123	1664	Standard

Given a base, and two positive integers in decimal (base 10), convert the two numbers to the new base, add them, and display their sum in the new base.

Input

Three positive integers denoting the base and the two numbers, respectively. Input numbers will be integers between 0 and 65535. Bases will be between 2 and 10 inclusive. Each case will be on a separate line. The end of input will be denoted by three zeros.

Output

An equation for the sum of the two numbers, in the new base.

Example

In this example, we add 10 and 3 in base 2, and we add 15 and 4 in base 3.
In base 2, 10 = 8 + 2 = 1*23 + 0*22 + 1*21 + 0*20 and 3 = 2 + 1 = 1*21 + 1*20, so their base 2 equivalents are 1010 and 11, respectively. 10 + 3 = 13 = 1*23+1*22+0*21+1*20, so the base 2 equivalent of 13 is 1101.
In base 3, 15 = 9 + 6 = 1*32+2*31+0*30 and 4 = 3 + 1 = 1*31 + 1*30, so their base 2 equivalents are 120 and 11, respectively. 15 + 4 = 19 = 2*32 + 0*31 + 1*30, so the base 3 equivalent of 19 is 201.

Input

2 10 3
3 15 4
0 0 0

Output

1010 + 11 = 1101
120 + 11 = 201

#include<stdio.h>

int main()

{   void change(int ,int);

int a,b,n,sum;

scanf("%d %d %d",&n,&a,&b);

while(a||b||n)

{ sum=a+b;

change(a,n);

printf(" + ");

change(b,n);

printf(" = ");

change(sum,n);

printf("/n");

scanf("%d %d %d",&n,&a,&b);

}

return 0;

}

void change(int x,int n)

{  int stack[50]={0};

int i=1;

while(x)

{ stack[i]=x%n;

i++;

x=x/n;

}

for(i--;i>=1;i--)

printf("%d",stack[i]);

}