手工验算一下吧，又不难的!
#include
<stdio.h>
//判断能否被合理的分配
int
divide(int
n,int
m)
{
if(n/5==0
||
n%5!=1)
{//不足5个或不能分5份多1个，分配失败
return
0
}
if(m==1)
{//分到最后一个猴子，说明能分配成功
return
1
}
return
divide(n-n/5-1,m-1)
}
main()
{
int
n//桃子数量
for(n=1n++)
{
if(divide(n,5))
{//判断能否被合理的分配
printf("%d\n",n)
break
}
}
}
#include <stdlib.h>#include<stdio.h>
#define N 10
#define M 20
struct monkey
{
int id
int tao
}
struct queue
{
struct monkey content[N]
struct monkey* head
struct monkey* tail
int length
}
int queue_init(struct queue** q)
{
(*q) = (struct queue*)malloc(sizeof(struct queue))
(*q)->head = &(*q)->content[0]
(*q)->tail = &(*q)->content[0]
(*q)->length = 0
return 0
}
int queue_in(struct queue* q,struct monkey in)
{
if(q->length == 0)
{
*(q->tail) = in
q->length++
}
else
{
q->tail++
*(q->tail) = in
q->length++
}
return 0
}
int queue_out(struct queue* q, struct monkey* out)
{
int i = 0
*(out) = *(q->head)
for(i <q->length-1i++)
q->content[i] = q->content[i+1]
q->tail--
q->length--
return 0
}
//队首到队尾
int queue_a(struct queue* q)
{
struct monkey temp
queue_out(q, &temp)
queue_in(q,temp)
return 0
}
int main()
{
struct monkey data[N]
struct queue* q = NULL
struct monkey temp
int i
int kuang = 0
for(i=0i<Ni++)
{
data[i].id = i+1
data[i].tao = 0
}
queue_init(&q)
for(i=0i<Ni++)
queue_in(q, data[i])
for(i=0i<Ni++)
{
printf("%d\t%d\n",q->content[i].id,q->content[i].tao)
}
printf("leng=%d\n",q->length)
printf("head %d\t tail %d\n",q->head->id,q->tail->id)
i = 0
while(q->length != 0)
{
if(kuang == 0)
{
i++
kuang += i
}
//队首猴取桃
if(q->head->tao + kuang <M)
{
q->head->tao += kuang
kuang = 0
queue_a(q)
}
else
{
kuang -= M - q->head->tao
q->head->tao += M - q->head->tao
queue_out(q, &temp)
printf("%d\t%d\n",temp.id,temp.tao)
}
}
return 0
}
死循环了！if(m-1>0&&(m-1)%5==0&&(m-1)/5!=0)//判断桃子能否合题分配
这句里，之前 m=1，这里始终不能成立，if 不运行，然后就一直循环。。。。
