题目
1058 A+B in Hogwarts(20 分)
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
题解
#include<iostream>
int main()
{
long a[3]={},b[3]={},c[3]={};
scanf("%ld.%ld.%ld %ld.%ld.%ld",&a[0],&b[0],&c[0],&a[1],&b[1],&c[1]);
if(c[0]+c[1]>=29)
{
c[2]=(c[0]+c[1])%29;
b[2]++;
}else{
c[2]=c[0]+c[1];
}
if(b[0]+b[1]+b[2]>=17)
{
b[2]+=(b[0]+b[1])-17;
a[2]++;
}else{
b[2]+=b[0]+b[1];
}
a[2]+=a[0]+a[1];
printf("%ld.%ld.%ld\n",a[2],b[2],c[2]);
return 0;
}
这算是一种最朴素的解法了。记得数组要初始化。还有,防止大数溢出用long,甚至long long。
#include <iostream>
using namespace std;
int main() {
long long a, b, c, d, e, f;
scanf("%lld.%lld.%lld %lld.%lld.%lld", &a, &b, &c, &d, &e, &f);
long long num = c + b * 29 + a * 29 * 17 + f + e * 29 + d * 29 * 17;
long long g = num / (17 * 29);
num = num % (17 * 29);
printf("%lld.%lld.%lld", g, num / 29, num % 29);
return 0;
}
柳神偏爱将总的数值算出来,然后用/两个进制的乘积算出从右往左第三位的值,(同理/一个进制值=从右往左第二位的值),用%(两个进制值相乘)算出后两位的数值,再得最后一位的值。
总结就是“/”1位进制值得倒数第二位,“%”1位进制值得倒数一位