void main()
{
/*0表示关灯，1表示开灯*/
int n,k,a[101],i,b,c,d
printf("输入人数")
scanf("%d",&k)
printf("输入灯的数目")
scanf("%d",&n)
for(i=1i<101i++)
{
a[i]=0
}
for(b=1b<=kb++) //这里不能是k++啊！死循环了
{
for(c=1c<=nc++)
{
if(c%b==0)
{
if(a[c]==0)
{
a[c]=1
}else
{
a[c]=0
}
}
}
}
for(d=1d<=nd++)
if(a[d]==1){
printf("%d ",d) //加上空格，看着清楚些
}
}
既然只要最后一个，何必全部计算。。。
#include<stdio.h>int main() {
int m, n
while (scanf("%d%d", &n, &m) != EOF) {
int isOn = 0
do
if (n % m == 0)
isOn = !isOn
while (--m)
printf("%s\n", isOn ? "on" : "off")
}
return 0
}
楼上那位同学是不知道有种东西叫oj么。。。
#include <stdio.h>#define n 100
void main(void)
{
int i, j, a[n]
for(i = 0i <ni++) a[i] = 0// 初始化为全0
for(i = 1i <= ni++)
for(j = i j <= nj = j + i)
a[j - 1] += 1
//100人对100灯，都操作完毕，下面是显示
for(i = 0i <ni++) {
if((i % 8) == 0) printf("\n")
printf(" L%2d: %d, ", i + 1, a[i] % 2)
}
printf("\n")
}
/*
运行结果显示如下：
L 1: 1, L 2: 0, L 3: 0, L 4: 1, L 5: 0, L 6: 0, L 7: 0, L 8: 0,
L 9: 1, L10: 0, L11: 0, L12: 0, L13: 0, L14: 0, L15: 0, L16: 1,
L17: 0, L18: 0, L19: 0, L20: 0, L21: 0, L22: 0, L23: 0, L24: 0,
L25: 1, L26: 0, L27: 0, L28: 0, L29: 0, L30: 0, L31: 0, L32: 0,
L33: 0, L34: 0, L35: 0, L36: 1, L37: 0, L38: 0, L39: 0, L40: 0,
L41: 0, L42: 0, L43: 0, L44: 0, L45: 0, L46: 0, L47: 0, L48: 0,
L49: 1, L50: 0, L51: 0, L52: 0, L53: 0, L54: 0, L55: 0, L56: 0,
L57: 0, L58: 0, L59: 0, L60: 0, L61: 0, L62: 0, L63: 0, L64: 1,
L65: 0, L66: 0, L67: 0, L68: 0, L69: 0, L70: 0, L71: 0, L72: 0,
L73: 0, L74: 0, L75: 0, L76: 0, L77: 0, L78: 0, L79: 0, L80: 0,
L81: 1, L82: 0, L83: 0, L84: 0, L85: 0, L86: 0, L87: 0, L88: 0,
L89: 0, L90: 0, L91: 0, L92: 0, L93: 0, L94: 0, L95: 0, L96: 0,
L97: 0, L98: 0, L99: 0, L100: 1,
Press any key to continue
*/
