标签:
题目算法不难,难的是读懂题意,意思是从1到N的数字转换成罗马数字,然后统计所有数字中的各种字母出现的次数
对于每个数,用贪心的方法转换为罗马数字,然后统计就好了
/* ID: modengd1 PROG: preface LANG: C++ */ #include <iostream> #include <stdio.h> #include <memory.h> #include <string> #include <cstring> using namespace std; string St[15]={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"}; int value[15]={1000,900,500,400, 100,90,50,40, 10, 9,5,4,1}; int counter[7]; char output[10]={‘I‘,‘V‘,‘X‘,‘L‘,‘C‘,‘D‘,‘M‘}; void Count(string S) { for(int i=0;i<S.size();i++) { switch(S[i]) { case ‘I‘: counter[0]++;break; case ‘V‘: counter[1]++;break; case‘X‘: counter[2]++;break; case‘L‘: counter[3]++;break; case‘C‘: counter[4]++;break; case‘D‘: counter[5]++;break; case‘M‘: counter[6]++;break; } } } string change(int x) { string s; while(x>0) { for(int j=0;j<15;j++) { if(x>=value[j]) { x-=value[j]; for(int k=0;k<St[j].size();k++) s.push_back(St[j][k]); break; } } } return s; } int main() { freopen("preface.in","r",stdin); freopen("preface.out","w",stdout); int N; memset(counter,0,sizeof(counter)); scanf("%d",&N); for(int i=1;i<=N;i++) { Count(change(i)); } for(int i=0;i<7;i++) if(counter[i]) cout<<output[i]<<‘ ‘<<counter[i]<<endl; return 0; }
标签:
原文地址:http://www.cnblogs.com/modengdubai/p/4780599.html