1 #include
      
        2 #define N 60 3 int exchage(float n,float *a,int c,float *r); 4 void main() 5 { 6     float rmb[]={
    100,50,20,10,5,2,1,0.5,0.2,0.1}; 7     int n=sizeof(rmb)/sizeof(rmb[0]),k,i; 8     float change,r[N];; 9     printf("请输入要找的零钱数:");10     scanf("%f",&change);11     for(i=0;i
       
        =rmb[i])13             break;14     k=exchage(change,&rmb[i],n-i,r);15     if(k<=0)16         printf("找不开!\n");17     else18     {19         printf("找零钱的方案:%.2f=",change);20         if(r[0]>=1.0)21                 printf("%.0f",r[0]);22         else23                 printf("%.2f",r[0]);24         for(i=1;i
        
         =1.0)27                 printf("+%.0f",r[i]);28             else29                 printf("+%.2f",r[i]);30         }31         printf("\n");32     }33 }34 int exchage(float n,float *a,int c,float *r)35 {36     int m;37     if(n==0.0)                /*能分解，分解完成*/38         return 0;39     if(c==0)                /*不能分解*/40         return -1;41     if(n<*a)42         return exchage(n,a+1,c-1,r);    /*继续寻找合适的面值*/43     else44     {45         *r=*a;                    /*将零钱保存到r中*/46         m=exchage(n-*a,a,c,r+1);        /*继续分解剩下的零钱*/47         if(m>=0)48             return m+1;            /*返回找零的零钱张数*/49         return -1;50     }51 }