Preface Numbering
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 }
