A1058 A+B in Hogwarts (20 point(s))

不同进位数相加

1. 原文

原题

If you are a fan of Harry Potter, you would know the world of magic has its own currency system – as Hagrid explained it to Harry, “Seventeen silver Sickles to a Galleon and twenty-nine Knuts to a Sickle, it’s easy enough.” Your job is to write a program to compute A+B where A and B are given in the standard form of Galleon.Sickle.Knut (Galleon is an integer in [0,$10^7$], Sickle is an integer in [0, 17), and Knut is an integer in [0, 29)).

Input Specification:

Each input file contains one test case which occupies a line with A and B in the standard form, separated by one space.

output Specification:

For each test case you should output the sum of A and B in one line, with the same format as the input.

Sample Input:

3.2.1 10.16.27

Sample output:

14.1.28

2. 解析

不同进位 十位17 个位29

3. AC代码

#include<cstdio>
int main()
{
	long long g1,s1,k1,g2,s2,k2;
	scanf("%lld.%lld.%lld %lld.%lld.%lld",&g1,&s1,&k1,&g2,&s2,&k2);
	long long a=g1*17*29+s1*29+k1;
	long long b=g2*17*29+s2*29+k2;
	long long sum=a+b;
    printf("%lld.%lld.%lld\n",sum/(17*29),sum/29%17,sum%29);
	return 0;
}

数字可能溢出，应该用long long

Way2:

#include<cstdio>
int num1[3],num2[3];
int ans[3]={};
int main(){
	scanf("%d.%d.%d %d.%d.%d",
		&num1[0],&num1[1],&num1[2],
		&num2[0],&num2[1],&num2[2]);
	int temp = num1[2]+num2[2];
	ans[2]=temp%29;
	temp/=29;
	temp = num1[1]+num2[1]+temp;
	ans[1]=temp%17;
	temp/=17;
	ans[0]=temp+num1[0]+num2[0];
	printf("%d.%d.%d\n",ans[0],ans[1],ans[2]);
	return 0;
}