关于日本著名数学游戏专家中村义作教授提出这样一个问题：父亲将2520个桔子分给六个儿子。分完后父亲说：“老大将分给你的桔子的1/8给老二；老二拿到后连同原先的桔子分1/7给老三；老三拿到后连同原先的桔子分1/6给老四；老四拿到后连同原先的桔子分1/5给老五；老五拿到后连同原先的桔子分1/4给老六；老六拿到后连同原先的桔子分1/3给老大”。结果大家手中的桔子正好一样多。问六兄弟原来手中各有多少桔子？ 怎样用c语言编写程序，能编的，最好能给我注释一下，先谢了

这道题要从后往前算，2520的六分之一是420，（设老大到老六用a1--a6表示）

当老六分1/3给老大后，老大和老六都有420个桔子，即

a1+a6=840

a6-1/3a6=420 解得a6=630;a1=210

现在不管你从前往后推还是从后往前推都能做出来了。

程序我就不写了，我在网吧不好写，反正知道解题方法写程序不难的

#include<stdio.h><br />
#include<math.h><br />
#include "iostream"<br />
using namespace std;
int main()
    {
    int b1,b2,b3,b4,b5,b6;
    for( b1=8;b1<2520;b1=b1+8)
        {
        for(b2=1;b2<2520-b1;b2++)
            {
            int lao2=(b1/8+b2);
            if((2520/6==lao2*6/7)&&(0==lao2%7))
                for(b3=1;b3<2520-b1-b2;b3++)
                    {
                    int lao3=lao2/7+b3;
                    if((2520/6==lao3*5/6)&&(0==lao3%6))
                        for (b4=1;b4<2520-b1-b2-b3;b4++)
                            {
                            int lao4=lao3/6+b4;
                            if ((2520/6==lao4*4/5)&&(0==lao4%5))
                                for(b5=1;b5<2520-b1-b2-b3-b4;b5++)
                                    {
                                    int lao5=lao4/5+b5;
                                    if ((2520/6==lao5*3/4)&&(0==lao5%4))
                                        {
                                        b6=2520-b1-b2-b3-b4-b5;
                                        int lao6=lao5/4+b6;
                                        if((2520/6==lao6*2/3) &&(0==lao6%3))
                                            if (2520/6==(b1*7/8+lao6/3))
                                                {
                                                cout<<"***分2520个橘子***"<<endl;
                                                cout<<"老大:"<<b1<<endl;
                                                cout<<"老二:"<<b2<<endl;
                                                cout<<"老三:"<<b3<<endl;
                                                cout<<"老四:"<<b4<<endl;
                                                cout<<"老五:"<<b5<<endl;
                                                cout<<"老六:"<<b6<<endl;
                                                cout<<"******************"<<endl;
                                                }
                                        }
                                    }
                            }
                    }
            }
        }
   // system("pause");
    }