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PAT-A1058 A+B in Hogwarts (钱币转换)

2021-09-13 21:33:04  阅读:27  来源: 互联网

标签:PAT A1058 .% long 29 Sickle Hogwarts Galleon lld


A1058 A+B in Hogwarts (钱币转换)

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)).

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

结尾无空行

Sample Output:

14.1.28

题意:

输入两个Galleon.Sickle.Knut货币的值,将值相加,输出结果。

分析:

这道题说实话并不难,可我不愧为踩坑小能手,相信很多人这个题目没有拿到满分都是第二个测试点没有通过的吧,大家的问题应该是没有考虑到数值的溢出吧,所以把int改成long long 类型的就行了,但是我!哎,我是错在了c>=29和b>=17,我错在没有写这个等于号,题目中是不包含,改了之后成功AC!

易错点:数值溢出&数值范围

代码如下:

#include<cstdio>
int main(){
	long long a,b,c,d,e,f;//注意!
	scanf("%lld.%lld.%lld %lld.%lld.%lld",&a,&b,&c,&d,&e,&f);
	a+=d;
	b+=e;
	c+=f;
	if(c>=29){//注意!
		c%=29;
		b++;
	} 
	if(b>=17){
		b%=17;
		a++;
	}
	printf("%lld.%lld.%lld",a,b,c); 
	return 0;
}

标签:PAT,A1058,.%,long,29,Sickle,Hogwarts,Galleon,lld
来源: https://blog.csdn.net/weixin_46055626/article/details/120275921

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