求C语言大神帮忙做题!急!在线等!好的话再加分!

2025-03-16 01:03:25
推荐回答(3个)
回答1:

#include 
int main()
{
int sum=2;
int flag=-1;
int i,pos,end=1;
int count=1;

scanf("%d",&pos);

while(end)
{
if(flag>0)
{
i=1;
while(i {
if(count==pos)
{
printf("%d/%d\n",i,sum-i);
end=0;
break;
}
i++;
count++;
}
}
if(flag<0)
{
i=sum-1;
while(i>0)
{
if(count==pos)
{
printf("%d/%d\n",i,sum-i);
end=0;
break;
}
i--;
count++;
}
}
flag =-flag;
sum++;
}
getchar();getchar();
}

 

#include 

int main()
    {
    int t;
    long n,possible,sum,i,j;
    scanf("%d", &t);
    while (t--)
        {
        scanf("%d", &n);
        possible=sqrt(n*2.0);
        while (possible*(possible+1)>=2*n)
            {
            possible--;
            }
        sum=possible*(possible+1)/2;
        if (possible%2)
            {
            i=n-sum;
            j=possible+2-i;
            }
        else
            {
            j=n-sum;
            i=possible+2-j;
            }
        printf("TERM %d IS %d/%d\n", n, i, j);
        }
    }

回答2:

#include
main()
{
    int N,i,counter,n;
    while(scanf("%d",&n)!=EOF)
    {
        N=n;
        counter=1;
        i=1;
        while(n-i>0)      
        {
            n-=i;
            i++;
            counter++;
        }
        if(counter%2==0)
        {
             printf("TERM %d IS %d/%d\n",N,n,counter+1-n);
        }
        else
        {
            printf("TERM %d IS %d/%d\n",N,counter+1-n,n);
        }
    }
}

回答3:

这题挺简单的,给你个链接吧,没时间写代码:
http://galaas.blogbus.com/

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