Preface Numbering

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题意：给定页数N，求页码1~N（用罗马数字表示）中一共有多少个I、V、X、L、C、D、M

解题思路：

读入N
从1遍历至N，在遍历过程中统计各个字母的个数
统计中参考百度百科---罗马数字----组数规则（http://baike.baidu.com/link?url=Rt-mNHT_kqOLjmxWLE7mZFAl7iH4nmdYkWRbAzSPvM48GCiw6lRiVDUhAYV_XSng）

代码：

/*ID: zc.rene1LANG: CPROG: preface */#include<stdio.h>#include<stdlib.h>#include<string.h>enum letters{I=0, V, X, L, C, D, M};char letter_map[7] = {'I', 'V', 'X', 'L', 'C', 'D', 'M'};int result[7];void CalculateLetters(int num){    result[M] += (num/1000);    num %= 1000;    if (num >= 900)    {result[C]++;result[M]++;num -= 900;    }    if (num >= 500)    {result[D]++;num -= 500;    }    if (num >= 400)    {result[C]++;result[D]++;num -= 400;    }    result[C] += (num/100);    num %= 100;    if (num >= 90)    {result[X]++;result[C]++;num -= 90;    }    if (num >=50)    {result[L]++;num -= 50;    }    if (num >= 40)    {result[X]++;result[L]++;num -= 40;    }    result[X] += (num/10);    num %= 10;    if (num >= 9)    {result[I]++;result[X]++;num -= 9;    }    if (num >= 5)    {result[V]++;num -= 5;    }    if (num >= 4)    {result[I]++;result[V]++;num -= 4;    }    result[I] += num;}    int main(void){    FILE *fin, *fout;    int N, i;    fin = fopen("preface.in", "r");    fout = fopen("preface.out", "w");    fscanf(fin, "%d", &N);    memset(result, 0, 7*sizeof(int));    for (i=1; i<=N; i++)    {CalculateLetters(i);    }    for (i=I; i<=M; i++)    {if (result[i] != 0){    fprintf(fout, "%c %d\n", letter_map[i], result[i]);}    }    return 0;}