/*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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