section2.2.1freface

这道题考察的就是对于数组的遍历,首先构造一个二维数组,将所有基本罗马数字的值都保存在该数组中,然后进行查表,将传入函数的数进行位数分解,然后查表,即可得最终答案。代码如下。

#include<bits/stdc++.h>
using namespace std;
string pre[5][11]={{" "," "," "," "," "," "," "," "," "," "},
				{" ","I","II","III","IV","V","VI","VII","VIII","IX"},
				{" ","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"},
				{" ","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"},
				{" ","M","MM","MMM"}};
string change(int y){
	string str;
	str.append(pre[4][y/1000]);
	y=y-1000*(y/1000);
	str.append(pre[3][y/100]);
	y=y-100*(y/100);
	str.append(pre[2][y/10]);
	y=y-10*(y/10);
	str.append(pre[1][y]);
	return str;
}
int main (){
	freopen ("preface.in","r",stdin);
	freopen ("preface.out","w",stdout);
	string x;
	int pref[10];
	char prefa[10]={' ','I','V','X','L','C','D','M'};
	memset(pref,0,sizeof(pref));
	int N=0,y=0,i=0,j=0;
	cin>>N;
	for (i=1;i<=N;i++){
		x=change(i);
		for (j=0;j<x.length();j++){
			for (int k=1;k<=7;k++){
				if (x[j]==prefa[k]){
					pref[k]++;
					break;
				}
			}
		}
	}
	for (i=1;i<=7;i++){
		if (pref[i]!=0){
			cout<<prefa[i]<<" "<<pref[i]<<endl;
		}
	}
	return 0;
}

 

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