ACM习题 All your baseAll your base 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.InputThree positive integers denoting the base and the two num

ACM习题 All your baseAll your base 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.InputThree positive integers denoting the base and the two num
ACM习题 All your base
All your base
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

ACM习题 All your baseAll your base 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.InputThree positive integers denoting the base and the two num
#include
using namespace std;
int tran(int base,int n)
{
int a[50];
int i=0,j;
while(n!=0)
{
a[i]=n%base;
n=n/base;
i++;
}
for(j=i-1;j>=0;j--)
coutbase>>a>>b)
{
if(base==0&&a==0&&b==0)
break;
tran(base,a);
cout