poj2685——Numeral System新进制系统

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Numeral System
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2227 Accepted: 1713

Description

Prof. Hachioji has devised a new numeral system of integral numbers with four lowercase letters "m", "c", "x", "i" and with eight digits "2", "3", "4", "5", "6", "7", "8", "9". He doesn't use digit "0" nor digit "1" in this system. 

The letters "m", "c", "x" and "i" correspond to 1000, 100, 10 and 1, respectively, and the digits "2", ...,"9" correspond to 2, ..., 9, respectively. This system has nothing to do with the Roman numeral system. 

For example, character strings 

"5m2c3x4i", "m2c4i" and "5m2c3x" 
correspond to the integral numbers 5234 (=5*1000+2*100+3*10+4*1), 1204 (=1000+2*100+4*1), and 5230 (=5*1000+2*100+3*10), respectively. The parts of strings in the above example, "5m", "2c", "3x" and "4i" represent 5000 (=5*1000), 200 (=2*100), 30 (=3*10) and 4 (=4*1), respectively. 

Each of the letters "m", "c", "x" and "i" may be prefixed by one of the digits "2", "3", ..., "9". In that case, the prefix digit and the letter are regarded as a pair. A pair that consists of a prefix digit and a letter corresponds to an integer that is equal to the original value of the letter multiplied by the value of the prefix digit. 

For each letter "m", "c", "x" and "i", the number of its occurrence in a string is at most one. When it has a prefix digit, it should appear together with the prefix digit. The letters "m", "c", "x" and "i" must appear in this order, from left to right. Moreover, when a digit exists in a string, it should appear as the prefix digit of the following letter. Each letter may be omitted in a string, but the whole string must not be empty. A string made in this manner is called an MCXI-string. 

An MCXI-string corresponds to a positive integer that is the sum of the values of the letters and those of the pairs contained in it as mentioned above. The positive integer corresponding to an MCXI-string is called its MCXI-value. Moreover, given an integer from 1 to 9999, there is a unique MCXI-string whose MCXI-value is equal to the given integer. For example, the MCXI-value of an MCXI-string "m2c4i" is 1204 that is equal to 1000 + 2*100 + 4*1. There are no MCXI-strings but "m2c4i" that correspond to 1204. Note that strings "1m2c4i", "mcc4i", "m2c0x4i", and "2cm4i" are not valid MCXI-strings. The reasons are use of "1", multiple occurrences of "c", use of "0", and the wrong order of "c" and "m", respectively. 

Your job is to write a program for Prof. Hachioji that reads two MCXI-strings, computes the sum of their MCXI-values, and prints the MCXI-string corresponding to the result.

Input

The input is as follows. The first line contains a positive integer n (<= 500) that indicates the number of the following lines. The k+1 th line is the specification of the k th computation (k=1, ..., n). 


specification1 
specification2 
... 
specificationn 

Each specification is described in a line: 

MCXI-string1 MCXI-string2 
The two MCXI-strings are separated by a space. 

You may assume that the sum of the two MCXI-values of the two MCXI-strings in each specification is less than or equal to 9999.

Output

For each specification, your program should print an MCXI-string in a line. Its MCXI-value should be the sum of the two MCXI-values of the MCXI-strings in the specification. No other characters should appear in the output.

Sample Input

10xi x9ii 9ic2x2i 4c8x8im2ci 4m7c9x8i9c9x9i ii 9m9c9x8im ii mm9i i9m8c7xi c2x8i

Sample Output

3xx6cx5m9c9x9im9m9c9x9imimimx9m9c9x9i


整体思路:

先把数转化为十进制进行加减得出相应结果;

再将结果转换为新进制输出;

<span style="font-size:18px;">/** Filename:    code.cpp* Created:     2016-04-29* Author:        yunlong Wang  *[mail:17744454343@163.com]* Desciption:  Desciption*/#include <cstdio>#include <cstdlib>#include <iostream>#include <stack>#include <queue>#include <algorithm>#include <cstring>#include <cmath>#include <vector>#include <bitset>#include <list>#include <sstream>#include <set>#include <functional>using namespace std;#define INT_MAX 1 << 30#define MAX 100typedef long long ll;int n;char a[10],b[10];int ans;int change(char c){    int anss;    if(c == 'm')anss = 1000;    else if(c == 'c')anss = 100;    else if(c == 'x')anss = 10;    else if(c == 'i')anss = 1;    return anss;}int c_d(char d[100]){    int summ = 0;    for (int i = 0; i < strlen(d); i += 1)    {        if ((d[i] >= 50) && (d[i] <= 57))        {            int j = i+1;            int a = int(d[i]-48);   //cout << a << endl;            int b = change(d[j]);   //cout << b << endl;            summ += a*b;            //cout << summ << endl;            i++;        }        else        {            summ += change(d[i]);        }    }    return summ;}int main(){    scanf("%d",&n);    while (n--)    {        scanf("%s",a);        scanf("%s",b);        ans = c_d(a)+c_d(b);            //将a,b转换为十进制数                                        //以下为将十进制数转换输出        int j = 0,a1,k = 0;               char d[100] = {0};        a1 = ans;        while (a1)                      //求整数的位数        {            a1 /= 10;            j++;        }        int t1,t2,t,j1=j;        for (int l = 0; l < j; l += 1)  //求整数中每一位并进行转换        {                        t1 = ans%int(pow(10.0,j1));            t2 = pow(10.0,j1-1);            t = t1/t2;                          if(t == 1)            {                if(j1 == 4)      d[k] = 'm';                else if(j1 == 3) d[k] = 'c';                else if(j1 == 2) d[k] = 'x';                else if(j1 == 1) d[k] = 'i';            }            else if(t!=0)               //为0时相应mcxi字符不显示            {                int k1 = k+1;                if(j1 == 4)     {d[k] = (t+48);d[k1] = 'm';}                else if(j1 == 3){d[k] = (t+48);d[k1] = 'c';}                else if(j1 == 2){d[k] = (t+48);d[k1] = 'x';}                else if(j1 == 1){d[k] = (t+48);d[k1] = 'i';}                k++;            }            if(t!=0)k++;                //为0时不移位,否则逢d[k]遇0停止输出            j1--;        }        cout << d << endl;              }    return 0;}</span>



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