/*此程序用来实现发工资时如何准备足够的零钱的问题*/
/*2005-4-5 梁见斌*/
#include<stdio.h>
#include<stdlib.h>
int main(void)
{/*ShuLiang_GeRen[13]用来存储各种数额的钞票的数量*/
   int i, j, num=0, ShuLiang_GeRen[13]={0}, ShuLiang_ZongShu[13]={0};
   float danwei[13]={100,50,20,10,5,2,1,0.5,0.2,0.1,0.05,0.02,0.01};
   float money, sum=0;
   printf("\n输入第%d个人的工资（输入负数表示结束）:", ++num);
      scanf("%f",&money);
   while(money >= 0)/*循环输入，直到输入负值*/
   {
         for(j=0; j<13; j++)/*先把每个人的钞票数量弄好*/
          ShuLiang_GeRen[j]=0;        
       i=0;
       sum += money;
         while(i < 13 && money > 0)   /*按先付大钞票，再付小钞票的顺序把每种钞票的数量存入数组*/
      {
         ShuLiang_GeRen[i] = money/danwei[i];
         money -= ShuLiang_GeRen[i]*danwei[i];
         i++;
      }
      for(j=0; j<13; j++)/*把所有人的钞票数量加起来*/
          ShuLiang_ZongShu[j] += ShuLiang_GeRen[j];
         printf("\n输入第%d个人的工资（输入负数表示结束）:", ++num);
         scanf("%f",&money);
    }
   printf("\n总共有%d个人，你一共需要准备%f元，其中", num, sum);    
   for(i=0; i<13; i++)
      printf(" %d 张 %f 元的\n", ShuLiang_ZongShu[i], danwei[i]);
   system("pause");    
   return 0;
}

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