PAT甲级.1058. A+B in Hogwarts (20)

1058. A+B in Hogwarts (20)

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题目：

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, 107], Sickle is an integer in [0, 17), and Knut is an integer in [0, 29)).

输入格式：

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

输出格式：

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

输入样例：

3.2.1 10.16.27

输出样例：

14.1.28

题意：

计算输入的A.B的和，然后转换进制输出

PAT链接

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思路：

version1.0

计算出A+B的和直接输出

version2.0

设置int型变量carry存放每一位相加后的仅为

c[i] = (a[i] + b[i] +carry) % mod;
carry = (a[i] + b[i] + carry) / mod;

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代码：

version1.0

/**
* @tag     PAT_A_1058
* @authors R11happy (xushuai100@126.com)
* @date    2016-8-20 00:13-00:27
* @version 1.0
* @Language C++
* @Ranking  280/2992
* @function null
*/

#include <cstdio>
#include <cstdlib>
#include <cstring>

const int Galleon = 29*17;
const int Sickle = 29;

typedef long long LL;
int main(int argc, char const *argv[])
{
    LL a1, b1, c1;  //如果用int类型会导致测试点3错误
    LL a2, b2, c2;
    LL sum;
    scanf("%lld.%lld.%lld", &a1, &b1, &c1);
    scanf("%lld.%lld.%lld", &a2, &b2, &c2);
    sum = (a1*Galleon+b1*Sickle+c1) + (a2*Galleon+b2*Sickle+c2);
    printf("%d.%d.%d\n",int(sum / Galleon), int(sum % Galleon / Sickle), int(sum % Sickle) );
    return 0;
}

version2.0

/**
* @tag     PAT_A_1058
* @authors R11happy (xushuai100@126.com)
* @date    2016-8-20 00:13-00:27
* @version 1.0
* @Language C++
* @Ranking  280/2992
* @function null
*/

#include <cstdio>
#include <cstdlib>
#include <cstring>

int main(int argc, char const *argv[])
{
    int a[3], b[3], c[3];
    scanf("%d.%d.%d", &a[0], &a[1], &a[2]);
    scanf("%d.%d.%d", &b[0], &b[1], &b[2]);
    int carry = 0;  //进位
    c[2] = (a[2] + b[2]) % 29;
    carry = (a[2] + b[2]) / 29;
    c[1] = (a[1] + b[1] + carry) % 17;
    carry = (a[1] + b[1] + carry) / 17;
    c[0] = a[0] + b[0] + carry;
    printf("%d.%d.%d\n", c[0], c[1], c[2] );
    return 0;
}

c[i] = (a[i] + b[i] +carry) % mod;
carry = (a[i] + b[i] + carry) / mod;

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收获：

1.int 类型可以保证8位以内都没问题，9位及9位以上要换long long
2.可以考虑version2.0加个carry的形式，这样位数再高似乎也没问题。