一、学习要点：
1.运用递归的思想，找出罗马字符串中最大的元素，对左边的处理是减去，对右边的处理是加上；递归头是字符串只剩下一个元素时，或者右边的下标小于左边的下标；
二、代码：

#include<stdlib.h>
#include<stdio.h>
int single_r2i(char ch)
{char s1[7] = { 'I','V','X','L','C','D','M' };int  s2[7]= { 1,5,10,50,100,500,1000 };for (int i = 0; i < 7; i++){if (s1[i] == ch) {return s2[i];}}
}
int find_maxindex(char* s, int l, int r) {int max = single_r2i(s[l]);int max_index=l;for (int i = l+1; i <= r; i++){if (single_r2i(s[i]) > max){max = single_r2i(s[i]);max_index = i;}}return max_index;
}
int r2i(char* s, int l, int r) {int max_index=0;if (l == r) {return single_r2i(s[l]);}if (l > r){return 0;}else {max_index = find_maxindex(s, l, r);return single_r2i(s[max_index]) + r2i(s, max_index + 1, r) - r2i(s, l, max_index - 1);}
}
int main()
{char s[7] = { 'M','C','M','X','C','I','X' };int result = r2i(s, 0, 6);printf("%d\n", result);system("pause");return 0;
}

三、程序运行结果：