USACO Preface Numbering 构造

一开始看到这道题目的时候，感觉好难

还要算出罗马的规则。

但是仔细一看，数据规模很小， n　只给到３５００

看完题目给出了几组样例之后就有感觉了

解题方法就是：

ｎ的每个十进制数　转换成相应的罗马数字，然后统计每个罗马数字出现的次数即可

还是一道简单的构造题。

(以下摘自https://www.byvoid.com/blog/usaco-221preface-numbering/)

转化有以下规则：

1、数较大部分在前，较小部分在后

2、表示10整倍数的字母(I X C M)最多可以累加三次

3、要累加4次的数应该将比该数的字母稍大的表示5整倍数或是10的整倍数的字母在后，累加的字母在前(例如IV XL CD CM)

了解以上规则后发现并不需要实际“转化”出罗马数字，而只用统计每个字母出现的次数。

My Source Code:

/*
ID: wushuai2
PROG: preface
LANG: C++
*/
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-7      ;
const int M = 660000         ;
const ll P = 10000000097ll   ;
const int INF = 0x3f3f3f3f   ;
const int MAX_N = 20         ;
const int MAXSIZE = 101000000;

int N, ans[10];
string op[5][10] = {
                    {},
                    {"", "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"}
};

void add(string str){
    int i, j;
    for(i = 0; i < str.length(); ++i){
        if(str[i] == ‘I‘)   ++ans[1];
        else if(str[i] == ‘V‘)  ++ans[2];
        else if(str[i] == ‘X‘)  ++ans[3];
        else if(str[i] == ‘L‘)  ++ans[4];
        else if(str[i] == ‘C‘)  ++ans[5];
        else if(str[i] == ‘D‘)  ++ans[6];
        else if(str[i] == ‘M‘)  ++ans[7];
    }
}

int main() {
    ofstream fout ("preface.out");
    ifstream fin ("preface.in");
    int i, j, k, t, n, s, c, w, q;
    fin >> N;
    for(i = 1; i <= N; ++i){
        int temp_g = i % 10;
        string temp = op[1][temp_g];
        add(temp);
        if(i < 10)  continue;
        int temp_s = (i / 10) % 10;
        temp = op[2][temp_s];
        add(temp);

        if(i < 100) continue;
        int temp_b = (i / 100) % 10;
        temp = op[3][temp_b];
        add(temp);

        if(i < 1000) continue;
        int temp_q = i / 1000;
        temp = op[4][temp_q];
        add(temp);
    }
    int flag = -1;
    for(i = 7; i >= 1; --i){
        if(ans[i]){
            flag = i;
            break;
        }
    }
    for(i = 1; i <= flag; ++i){
        if(i == 1){
            fout << "I" << ‘ ‘ << ans[i] << endl;
        } else if(i == 2){
            fout << "V" << ‘ ‘ << ans[i] << endl;
        } else if(i == 3){
            fout << "X" << ‘ ‘ << ans[i] << endl;
        } else if(i == 4){
            fout << "L" << ‘ ‘ << ans[i] << endl;
        } else if(i == 5){
            fout << "C" << ‘ ‘ << ans[i] << endl;
        } else if(i == 6){
            fout << "D" << ‘ ‘ << ans[i] << endl;
        } else if(i == 7){
            fout << "M" << ‘ ‘ << ans[i] << endl;
        }
    }

    fin.close();
    fout.close();
    return 0;
}