C语言编程班费问题,新人求助

2025-05-02 03:05:28
推荐回答(3个)
回答1:

#include
int main(void){
    int sum1,sum2,sum3,x;
    int s1,s2,s3;
    printf("请输入钱数:");
    scanf("%d",&x);
    sum1=x/6;
    sum2=x/5;
    sum3=x/4;  
    s1=x%6;
    s2=x%5;
    s3=x%4;
    if(sum1>sum2 && sum1>sum3 && s1==0){
        printf("买了%d支,6元的钢笔!",sum1);
    }
    else if(sum2>sum1 && sum2>sum3 && s2==0){
        printf("买了%d支,5元的钢笔!",sum2);
    }
    else if(sum3>sum1 && sum3>sum2 && s3==0){
        printf("买了%d支,4元的钢笔!",sum3);
    }
    else{
        printf("No Answer");
    }
    return 0;
}

回答2:

int main()
{
    int i,j,n;
    bool isSolved=false;
    scanf("%d",&n);
    for(i=0;i<=n/4;i++)
    {
        int tmp=(n-4*i)/5;
        if(tmp) for(j=0;j<=tmp;j++)
        {
             if(0==(n-4*i-5*j)%6)
             {
                 printf("%d %d %d\n",(n-4*i-5*j)/6,j,i);
                 isSolved=true;
             }
        }
    }
    if(!isSolved) printf("No Answer\n");
    return 0;
}

回答3:

#include
#include
int main()
{
int x,i,j,k,m=0,n[1000]={0},h=0,p;
int a[1000]={0},b[1000]={0},c[1000]={0};
scanf("%d",&x);
p=x/4;
for(i=1;ifor(j=1;jif((x-6*i-5*j)%4==0&&(x-6*i-5*j)!=0)
{
k=(x-6*i-5*j)/4;
n[m]=i+j+k;
a[m]=i;
b[m]=j;
c[m]=k;
m++;
}
if(m==0)
printf("No Answer\n");
else
{
for(i=0;iif(n[h]{h=i;}
printf("%d %d %d\n",a[h],b[h],c[h]);
}
return 0;
}