某品牌巧克力使用500克原料可制作55小块巧克力，请编程实现：输入原料重量（以千克为单位），计算出制作巧克力的块数（四舍五入）。然后对这些巧克力进行分包，小盒放11块，大盒放24块，问各分装多少大盒多少小盒剩余巧克力块数最少。

#include<stdio.h>
#include<math.h>
int main()
{
    int a,b,n,i,min;
    float m;
    printf("please input the weight(/kg):\n");
    scanf("%f",&m);
    min=i;
    for(a=0,b=0;i<=n;i--)
    {
        n=(int)(m/0.5*55+0.5);
        i=n-(11*a+24*b);
        if(i<min)
        {
            min=i;
            printf("%d %d\n",a,b);
        }
    }
    return 0;
}

#include<stdio.h>

#define PIECE_PER_KG    110.0        //每KG可生产的巧克力块数 = 1000 * 55 / 500
#define BIGBOX            24
#define SMALLBOX        11

int main(void)
{
    int n, big_rem, small_rem, big, small, temp, rem ;
    float m;
    printf("please input the weight(/kg):");
    scanf("%f",&m);
    n = (int)(PIECE_PER_KG * m) ;
    rem = n ;

    for ( big = 0; big * BIGBOX <= n; big++ ) {
        for ( small = 0; ( small * SMALLBOX + big * BIGBOX ) <= n; small++ ) {
            temp = n - small * SMALLBOX - big * BIGBOX ;
            if ( rem > temp ) {
                rem = temp ;
                big_rem = big ;
                small_rem = small ;
            }
        }
    }
    //printf ( "巧克力共有%d块，大盒装24块，小盒装11块。\n", n ) ;
    printf ( "大盒数量：%d个，小盒数量：%d个，剩余巧克力：%d块。\n", big_rem, small_rem, rem ) ;
    return 0;
}

我只能想到2楼的方法~

不会   伤心了