1 //递归函数整理 2 long fun(int n) 3 { 4     long s; 5     if(n==1||n==2) 6         s=2; 7     else 8         s=n-fun(n-1); 9     10     return s;11 }12 13 int main(int argc, const char * argv[])14 {15     printf("%ld",fun(6));16     /*17      = 6 - (5 - fun(4))18      = 6 - (5 - (4 - fun(3)))19      = 6 - (5 - (4 - (3 - fun(2))))20      = 6 - (5 - (4 - (3 - 2)))21      = 6 - (5 -4 + 3 - 2)22      = 423      */24     return 0;25 }