标签：ott using one har 直接 简单 stand ace tee
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 10.16.27Sample Output:
14.1.28
思路
加法，注意进位就好。
注：
1.这道题看着很简单，然而用cin输入的是两个字符串，还得将字符串分割处理，然后转换成int，有点坑。
2.直接用scanf("%d.%d.%d",&x,&y,&z)读数据就不用处理字符串 =_=!。
代码
#include<iostream> #include<vector> #include<string> using namespace std; int main() { vector<int> A(3),B(3); string a,b; while(cin >> a >> b) { int index = 0; //handle A for(int i = 0,j = 0;i <a.size();i++) { string tmp; if(a[i] == ‘.‘) { tmp = a.substr(j,i - j); j = i + 1; A[index++] = stoi(tmp); } if(index == 2) { tmp = a.substr(j,a.size() - j); A[index++] = stoi(tmp); } } //Handle B index = 0; for(int i = 0,j = 0;i <b.size();i++) { string tmp; if(b[i] == ‘.‘) { tmp = b.substr(j,i - j); j = i + 1; B[index++] = stoi(tmp); } if(index == 2) { tmp = b.substr(j,b.size() - j); B[index++] = stoi(tmp); } } int add = 0,sum = 0; sum = (A[2] + B[2]); add = sum / 29; A[2] = sum % 29; sum = A[1] + B[1] + add; add = sum/17; A[1] = sum % 17; A[0] = A[0] + B[0] + add; cout << A[0] << "." <<A[1] << "." << A[2]; } }
标签：ott using one har 直接 简单 stand ace tee
原文地址：http://www.cnblogs.com/0kk470/p/7624723.html
