1058. A+B in Hogwarts (20)
时间限制
50 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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
Sample Output:
14.1.28
分别相加进位
利用了stringstream
代码如下很low
#include<iostream>
#include<string>
#include<sstream>
using namespace std;
long change(string s){
stringstream ss;
long value;
ss<<s;
ss>>value;
return value;
}
int main(){
string a,b;
cin>>a>>b;
int index=a.find(".");
string a1=a.substr(0,index);
a.erase(0,index+1);
index=a.find(".");
string a2=a.substr(0,index);
a.erase(0,index+1);
string a3=a;
index=b.find(".");
string b1=b.substr(0,index);
b.erase(0,index+1);
index=b.find(".");
string b2=b.substr(0,index);
b.erase(0,index+1);
string b3=b;
long a11,a21,a31,b11,b21,b31,s1=0,s2=0,s3=0;
a11=change(a1);
a21=change(a2);
a31=change(a3);
b11=change(b1);
b21=change(b2);
b31=change(b3);
if((a31+b31)>=29){
s3=a31+b31-29;
s2++;
}else{
s3=a31+b31;
}
if((a21+b21+s2)>=17){
s2=s2+a21+b21-17;
s1++;
}else{
s2+=a21+b21;
}
s1=s1+a11+b11;
cout<<s1<<"."<<s2<<"."<<s3<<endl;
return 0;
}
流弊代码如下原文地址
完美利用了进制转换
#include<iostream>
using namespace std;
int main(){
long long a,b,c,e,f,g;
scanf("%lld.%lld.%lld %lld.%lld.%lld",&a,&b,&c,&e,&f,&g);
long long sum=a*29*17+b*29+c+e*29*17+f*29+g;
long long first=sum/(29*17);
sum=sum%(29*17);
printf("%lld.%lld.%lld",first,sum/29,sum%29);
return 0;
}
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