C语言 输入多个数相加或相减例如：
输入：2+4+2+10
输出结果：18
输入：11-2＋9+10+27
输出结果：55
输入：33－11-3
输出结果：19
求大神帮忙一下，有人说是这样的步骤，可是我不知道怎么写：1）读入字符串（2）提取运算数（3）提取运算符（4）计算

步骤清晰明了，读入字符串有函数，提取运算数和运算符就是找到字符串中的所有运算符(不要说这个不知道怎么找)，其左右相应位置就是运算数(需自己字符串转换为数字)了。

#include <stdio.h>

int foo( const char* s )
{
    int sum = 0;
    for( int i=0,v,j; sscanf(s+i,"%d%n",&v,&j)==1; i+=j )
        sum += v;
    return sum;
}

int main()
{
    printf( "%d\n", foo("2+4+2+10") );
    printf( "%d\n", foo("11-2+9+10+27") );
    printf( "%d\n", foo("33-11-3") );

    return 0;
}

#include <stdio.h>
int main()
{
   int sum = 0, times = 0, a, a1 = 0;
   printf("Please input a: (0 for add, others for subtract): \n");
   scanf("%d", &a);
   if(a == 0)
   {
      printf("Please input the NO. of the numbers: \n");
      scanf("%d", &times);
      for(int i = 0; i < times; i++)
      {
         scanf("%d", &a1);
         if(i == 0)
         {
            sum = i;
         };
         else
         {
            sum += i;
         };
      };
      printf("the result: %d\n", sum);
   };
   else
   {
      printf("Please input the NO. of the numbers: \n");
      scanf("%d", &times);
      for(int i = 0; i < times; i++)
      {
         scanf("%d", &a1);
         if(i == 0)
         {
            sum = i;
         };
         else
         {
            sum -= i;
         };
      };
      printf("the result: %d\n", sum);
   };
   return 0;
};

this program is so stupid after second time i seeing it. how I make it out ...

#include <stdio.h>

int foo( const char* s )
{
    int sum = 0;
    for( int i=0,v,j; sscanf(s+i,"%d%n",&v,&j)==1; i+=j )
        sum += v;
    return sum;
}

int main()
{
    char str[100];
    gets(str);
    printf( "%d\n", foo(str) );

    return 0;
}//3楼的源代码，我加了个读取字符串的两行。
3楼大神！！学习了！！！