又一版 A+BTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 15344 Accepted Submission(s): 5867
Problem Description
输入两个不超过整型定义的非负10进制整数A和B(<=2 31-1),输出A+B的m (1 < m <10)进制数。
Input
输入格式:测试输入包含若干测试用例。每个测试用例占一行,给出m和A,B的值。 当m为0时输入结束。
Output
输出格式:每个测试用例的输出占一行,输出A+B的m进制数。
Sample Input
8 1300 482 1 70
Sample Output
25041000
//注意得有n=0,m=0的特判(在这块WA了4次)
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define IN __int64
#define ull unsigned long long
#define ll long long
#define N 10010
#define M 1000000007
using namespace std;
int a[N];
int main()
{
int k;
while(scanf("%d",&k),k)
{
int i,j;
IN n,m;
scanf("%I64d%I64d",&n,&m);
if(n==0&&m==0)
{
printf("0\n");
continue;
}
memset(a,0,sizeof(a));
IN sum=n+m;
int kk=0;
while(sum)
{
a[kk++]=sum%k;
sum/=k;
}
for(j=kk-1;j>=0;j--)
printf("%d",a[j]);
printf("\n");
}
return 0;
}
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