45.將真分?jǐn)?shù)分解為埃及分?jǐn)?shù)
     分子為1 的分?jǐn)?shù)稱為埃及分?jǐn)?shù)，現(xiàn)輸入一個(gè)真分?jǐn)?shù)，請將該分?jǐn)?shù)分解為埃及分?jǐn)?shù)。
     如：8/11=1/2+1/5+1/55+1/110。
    *問題分析與算法設(shè)計(jì)
     若真分?jǐn)?shù)的分子a能整除分母b，則真分?jǐn)?shù)經(jīng)過化簡就可以得到埃及分?jǐn)?shù)，若真分?jǐn)?shù)的分子不能整除分母，則可以從原來的分?jǐn)?shù)中分解出一個(gè)分母為b/a+1的埃及分?jǐn)?shù)。用這種方法將剩余部分反復(fù)分解，最后可得到結(jié)果。
    *程序與程序注釋
    #include
    void main()
    {
     long int a,b,c;
     printf("Please enter a optional fraction(a/b):");
     scanf("%ld/%ld",&a,&b); /*輸入分子a和分母b*/
     printf("It can be decomposed to:");
     while(1)
     {
     if(b%a) /*若分子不能整除分母*/
     c=b/a+1; /*則分解出一個(gè)分母為b/a+1的埃及分?jǐn)?shù)*/
     else{ c=b/a; a=1;} /*否則，輸出化簡后的真分?jǐn)?shù)(埃及分?jǐn)?shù))*/
     if(a==1)
     {
     printf("1/%ld\n",c);
     break; /*a為1標(biāo)志結(jié)束*/
     }
     else
     printf("1/%ld + ",c);
     a=a*c-b; /*求出余數(shù)的分子*/
     b=b*c; /*求出余數(shù)的分母*/
     if(a==3) /*若余數(shù)為3，輸出最后兩個(gè)埃及分?jǐn)?shù)*/
     { printf("1/%ld + 1/%ld\n",b/2,b); break;}
     }
    }
    *運(yùn)行結(jié)果
     1. Please enter a optional fraction (a/b): 1/6
     It can be decomposed to: 1/6
     2. Please enter a optional fraction (a/b): 20/33
     It can be decomposed to: 1/2+1/10+1/165
     3. Please enter a optional fraction (a/b): 10/89
     It can be decomposed to: 1/9+1/801
     4. Please enter a optional fraction (a/b): 19/99
     It can be decomposed to: 1/6+1/40+1/3960
     5. Please enter a optional fraction (a/b): 8/89
     It can be decomposed to: 1/11+1/957