这一题如何写成循环，很简单的题目
#include <stdio.h>
void sb(int a)
{
int b,c,d,e,f,g,h,i;
b=a/100; //求需要多少百
c=a%100; //求十个位
d=c/50; //求需要多少个50
e=(c%50)/20; //求需要多少个20
f=((c%50)%20)/10; //求需要多少个10
g=(((c%50)%20)%10)/5; //求需要多少个5
h=((((c%50)%20)%10)%5)/2; //求需要多少个2
i=(((((c%50)%20)%10)%5)%2)/1; //求需要多少个1
printf( "%d个一百块\n%d个五十块\n%d个二十块\n%d个十块\n%d个五块\n%d个二块\n%d个一块\n ",b,d,e,f,g,h,i );


}

void main()
{
int a;
printf( "请输入你拥有多少钱 ");
scanf( "%d ",&a);
sb(a);

}

[解决办法]
unsigned int bill[] = {100,50,20,10,5,2,1,0};

void getBillCount(unsigned int cash, unsigned int bill_value,unsigned int *pCount, unsigned int *pRemaining)
{
assert(bill_value);
*pCount = cash / bill_value;
*pRemaining = cash % bill_value
}

void main()
{
unsigned int total_value;
printf( "请输入你拥有多少钱 ");
scanf( "%d ", &total_value); //输入数据需查错

unsigned int remaining = total_value;

for (int i = 0; ; ++i) //no end condition, will break out
{
unsigned int nCount = 0;
if (bill[i] == 0)
{
break; //we are end of here
}
getBillCount(remaining, bill[i], &nCount, &remaining);

printf( "%d个：%d 元\r\n ",nCount, bill[i]);
}
}

Notepad写的,可能有Bug.

另外题外话,这个算法如果是为了得到最少纸张数不是广适的,比如有25块钱的面值的钞票的话,就不对了. (40块钱的情况). 所以贪婪算法不一定得到最好结果.
[解决办法]
把这个函数换成这个:

void getBillCount(unsigned int cash, unsigned int bill_value,unsigned int *pCount, unsigned int *pRemaining)
{
// assert(bill_value);
*pCount = cash / bill_value;
*pRemaining = cash % bill_value;
}