USACO Section 2.2 Preface Numbering - 题目只是自始自终没太懂..-优快云博客

本文通过C++实现了一个程序，该程序统计从1到给定整数N内所有罗马数字中各个符号出现的次数。通过对每个罗马数字位的单独处理并进行组合，实现了对I、V、X、L、C、D、M等符号的有效计数。

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题目真心没看懂...后来是看了翻译都没懂....最后看了别人的分析..说每位单独的看再组合就ok了...但我也没懂..但真的很乱了...好吧..我就按那个思路做了一次码农...

Program:

/* ID: zzyzzy12 LANG: C++ TASK: preface */ #include<iostream> #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #include<queue> using namespace std; int N,ans[10]; string a[4][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"}; int turn(char c) { if (c=='I') return 1; if (c=='V') return 2; if (c=='X') return 3; if (c=='L') return 4; if (c=='C') return 5; if (c=='D') return 6; return 7; } void make(int n) { string s=""; int i,len=0; while (n>0) { s+=a[len][n%10]; n/=10; len++; } len=s.length(); for (i=0;i<len;i++) ans[turn(s[i])]++; } int main() { freopen("preface.in","r",stdin); freopen("preface.out","w",stdout); int i; memset(ans,0,sizeof(ans)); scanf("%d",&N); for (i=1;i<=N;i++) make(i); if (ans[1]) printf("I %d\n",ans[1]); if (ans[2]) printf("V %d\n",ans[2]); if (ans[3]) printf("X %d\n",ans[3]); if (ans[4]) printf("L %d\n",ans[4]); if (ans[5]) printf("C %d\n",ans[5]); if (ans[6]) printf("D %d\n",ans[6]); if (ans[7]) printf("M %d\n",ans[7]); return 0; }