百钱买百鸡的问题算是一套非常经典的不定方程的问题，题目很简单：公鸡5文钱一只，母鸡3文钱一只，小鸡3只一文钱，

用100文钱买一百只鸡,其中公鸡，母鸡，小鸡都必须要有，问公鸡，母鸡，小鸡要买多少只刚好凑足100文钱。

~~~~~~~~~

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int i=0,j=0,k=0;
    for (i=1;i<20;i++){
        n=n++;
        for (j=1;j<33;j++){
            n=n++;
            k = 100-i-j;
            if (i*5+j*3+ k/3 == 100 && k%3 == 0){
                printf("the value of cock is:%d\n",i); 
                printf("the value of hen is:%d\n",j); 
                printf("the value of chicken is:%d\n",k); 
                printf("~~~~~~~~~~~~~~~~~~~~~~~~~~\n");
            }
        }
    }
   

 printf("the total of count is:%d\n",n);
system("pause");
return 0;
}

时间复杂度为O(N²),

优化一点如下，记数器从627变为310：

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int i=0,j=0,k=0,n=0;
    for (i=1;i<20;i++){
        n=n++;
        for (j=1;j<((100-5*i)/3);j++){
            n=n++;
            k = 100-i-j;
            if (i*5+j*3+ k/3 == 100 && k%3 == 0){
                printf("the value of cock is:%d\n",i); 
                printf("the value of hen is:%d\n",j); 
                printf("the value of chicken is:%d\n",k); 
                printf("~~~~~~~~~~~~~~~~~~~~~~~~~~\n");
            }
        }
    }
   

 printf("the total of count is:%d\n",n);
system("pause");
return 0;
}

如果要成为O（N），则先用方程式多推导，这才是算法的精华。。

先在头脑中过滤，再交给CPU，内存去实施~~

抄其它人的算法如下：

              for (int k = 1; k <= 3; k++)
                 x = 4 * k;
                 y = 25 - 7 * k;
                 z = 75 + 3 * k;

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