zoj 1205 二十进制数相加

最新推荐文章于 2020-09-14 20:29:12 发布

转载最新推荐文章于 2020-09-14 20:29:12 发布 · 226 阅读

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原文链接：http://www.cnblogs.com/laipDIDI/articles/2042894.html

本文介绍了一个有趣的编程挑战，源于火星居民的数学竞赛。任务是在20进制下计算两个100位数的和，使用C++实现。文章提供了完整的代码示例，展示了如何将字符型数字转换为整数进行计算，并将结果转换回字符输出。

Martian Addition

Time Limit: 1 Second Memory Limit: 32768 KB

In the 22nd Century, scientists have discovered intelligent residents live on the Mars. Martians are very fond of mathematics. Every year, they would hold an Arithmetic Contest on Mars (ACM). The task of the contest is to calculate the sum of two 100-digit numbers, and the winner is the one who uses least time. This year they also invite people on Earth to join the contest.
As the only delegate of Earth, you're sent to Mars to demonstrate the power of mankind. Fortunately you have taken your laptop computer with you which can help you do the job quickly. Now the remaining problem is only to write a short program to calculate the sum of 2 given numbers. However, before you begin to program, you remember that the Martians use a 20-based number system as they usually have 20 fingers.

Input:
You're given several pairs of Martian numbers, each number on a line.
Martian number consists of digits from 0 to 9, and lower case letters from a to j (lower case letters starting from a to present 10, 11, ..., 19).
The length of the given number is never greater than 100.

Output:
For each pair of numbers, write the sum of the 2 numbers in a single line.

Sample Input:

1234567890 abcdefghij 99999jjjjj 9999900001

Sample Output:

bdfi02467j iiiij00000

ZOJ <wbr>1205 <wbr>Martian <wbr>Addition

#include < iostream >
#include < map >
#include < algorithm >
#include < string >
using namespace std;

int main()
{
map < char , int > char_int;
for ( int i = 0 ;i < 10 ;i ++ ){
char_int[ ' 0 ' + i] = i;
}
for ( int i = 0 ; i < 10 ;i ++ ){
char ch = ' a ' ;
char_int[ch + i] = i + 10 ;
}

map < int , char > int_char;
for ( int i = 0 ;i < 10 ;i ++ ){
int_char[i] = ' 0 ' + i;
}
for ( int i = 0 ; i < 10 ;i ++ ){
char ch = ' a ' ;
int_char[i + 10 ] = ch + i;
}

string num1, num2 ,num3;
while (cin >> num1 >> num2){
reverse(num1.begin(),num1.end());
reverse(num2.begin(),num2.end());
int up = 0 ;
num3 = "" ;
string str;

if (num1.size() < num2.size()){
str = num1;
num1 = num2;
num2 = str ;
}

for ( int i = 0 ;i < num2.size();i ++ ){
if (char_int[ num1[i] ] + char_int[num2[i]] + up >= 20 ){

num3 = int_char[char_int[ num1[i] ] + char_int[num2[i]] + up - 20 ] + num3;
up = 1 ;
}
else {

num3 = int_char[char_int[ num1[i] ] + char_int[num2[i]] + up] + num3;
up = 0 ;
}
}
if (up == 1 ){
for ( int i = num2.size();i < num1.size();i ++ ){
if (char_int[num1[i] ] + up >= 20 ){

num3 = int_char[char_int[num1[i]] + up - 20 ] + num3;
up = 1 ;
}
else {

num3 = int_char[char_int[ num1[i] ]] + num3;
up = 0 ;
}
}
}
else {
for ( int i = num2.size();i < num1.size();i ++ ){
num3 = int_char[char_int[ num1[i] ]] + num3;
}
}
if (up == 1 ){
num3 = ' 1 ' + num3;
}
cout << num3 << endl;
}
return 0 ;
}

转载于:https://www.cnblogs.com/laipDIDI/articles/2042894.html