Preface Numbering

链接

分析：先打表需要用到的罗马数字，然后暴力转换，最后统计一下即可

 1 /*
 2     PROB:preface
 3     ID:wanghan
 4     LANG:C++
 5 */
 6 #include "iostream"
 7 #include "cstdio"
 8 #include "cstring"
 9 #include "string"
10 #include "map"
11 using namespace std;
12 const int maxn=4000;
13 char s[]={‘I‘,‘V‘,‘X‘,‘L‘,‘C‘,‘D‘,‘M‘};
14 int  h[]={1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100,200,300,400,500,600,700,800,900,1000,2000,3000};
15 int n;
16 struct Node{
17     int x;
18     string str;
19 };
20 Node p[maxn];
21 map<int,string> mp;
22 void init(){
23     mp[1]="I",mp[2]="II",mp[3]="III",mp[4]="IV",mp[5]="V",mp[6]="VI",mp[7]="VII",mp[8]="VIII",mp[9]="IX";
24     mp[10]="X",mp[20]="XX",mp[30]="XXX",mp[40]="XL",mp[50]="L",mp[60]="LX",mp[70]="LXX",mp[80]="LXXX",mp[90]="XC";
25     mp[100]="C",mp[200]="CC",mp[300]="CCC",mp[400]="CD",mp[500]="D",mp[600]="DC",mp[700]="DCC",mp[800]="DCCC",mp[900]="CM";
26     mp[1000]="M",mp[2000]="MM",mp[3000]="MMM";
27 }
28 string solve(int num){
29     string res="";
30     while(num){
31         int pos=-1;
32         for(int i=29;i>=0;i--){
33             if(num>=h[i]){
34                 pos=i; break;
35             }
36         }
37         int mod=num/h[pos];
38         for(int i=1;i<=mod;i++){
39             num-=h[pos];
40             res+=mp[h[pos]];
41         }
42         //num-=h[pos];
43         //res+=mp[h[pos]];
44     }
45     return res;
46 }
47 map<char,int>mr;
48 int main()
49 {
50     freopen("preface.in","r",stdin);
51     freopen("preface.out","w",stdout);
52     cin>>n;
53     init();
54     for(int i=1;i<=n;i++){
55         p[i].x=i;
56         p[i].str=solve(i);
57         //cout<<i<<": ";
58         //cout<<p[i].str<<endl;
59     }
60     for(int i=1;i<=n;i++){
61         for(int j=0;j<p[i].str.length();j++){
62             mr[p[i].str[j]]++;
63         }
64     }
65     for(int i=0;i<7;i++){
66         if(mr[s[i]]==0)  continue;
67         else{
68             cout<<s[i]<<" "<<mr[s[i]]<<endl;
69         }
70     }
71     return 0;
72 }