2.2.1 Preface Numbering

数字比较小，所以题目的难点就转到了，阿拉伯数字向罗马数字转化的过程了。

转化也不难。我直接手算了，值得注意的是8的写法VIII（不是IIX）。

整体来说不难。只要观察出每一位是相互独立的就行。

具体代码如下：

/*
ID: awsd1231
PROG: preface
LANG: C++
*/
#include<iostream>
#include<cstdio>
#include<vector>
#include<map>

using namespace std;

int n;
map<char, int> ans;
const char ansDir[7] = {‘I‘, ‘V‘, ‘X‘, ‘L‘, ‘C‘, ‘D‘, ‘M‘};

struct T {
	string one;
	string ten;
	string hun;
	string tho;
	T () { one = "*"; }
}num[10];

void init() {// IVX XLC CDM
	ans[‘I‘] = ans[‘V‘] = ans[‘X‘] = ans[‘L‘] = ans[‘C‘] = ans[‘D‘] = ans[‘M‘] = 0;
	num[1].one = "I"; num[1].ten = "X"; num[1].hun = "C"; num[1].tho = "M";
	num[2].one = "II"; num[2].ten = "XX"; num[2].hun = "CC"; num[2].tho = "MM";
	num[3].one = "III"; num[3].ten = "XXX"; num[3].hun = "CCC"; num[3].tho = "MMM";
	num[4].one = "IV"; num[4].ten = "XL"; num[4].hun = "CD";
	num[5].one = "V"; num[5].ten = "L"; num[5].hun = "D";
	num[6].one = "VI"; num[6].ten = "LX"; num[6].hun = "DC";
	num[7].one = "VII"; num[7].ten = "LXX"; num[7].hun = "DCC";
	num[8].one = "VIII"; num[8].ten = "LXXX"; num[8].hun = "DCCC";//!!!!8的正确写法
	num[9].one = "IX"; num[9].ten = "XC"; num[9].hun = "CM";
}



void count(int x) {
	string ansTmp;
	int idx = 1;
	while (x) {
		int tmp = x % 10;
		if (idx == 1) ansTmp += num[tmp].one;
		if (idx == 2) ansTmp += num[tmp].ten;
		if (idx == 3) ansTmp += num[tmp].hun;
		if (idx == 4) ansTmp += num[tmp].tho;
		x /= 10;
		++idx;
	}
	int len = ansTmp.length();
	for(int i = 0; i != len; ++i) {
		++ans[ansTmp[i]];
	}
}


int main () {
	freopen("preface.in", "r", stdin); 
	freopen("preface.out", "w", stdout);
	scanf("%d", &n);
	init();
	for (int i = 1; i != n + 1; ++i) {
		count(i);
	}
	for (int i = 0; i != 7; ++i) {
		if (ans[ansDir[i]]) cout << ansDir[i] << " " << ans[ansDir[i]] << endl;
	}
	return 0;
}

来自为知笔记(Wiz)