PAT 7-38 数列求和-加强版

 代码如下：

#include<stdio.h>
int main()
{
  int x,n;
  scanf("%d%d",&x,&n);
  if(n==0){printf("0");return 0;}
  int accum[99999]={0},i,j=0,temp=0,p=0;
  for(i=n;i>0;i--)
  {
    temp=i*x+p;
    p=temp/10;
    accum[j++]=temp%10;
  }
  if(p!=0)
  printf("%d",p);
  for(i=j-1;i>=0;i--)
  printf("%d",accum[i]);

}

　　参考代码：

#include<stdio.h>
int main()
{
  int x,n;
  scanf("%d%d",&x,&n);
  int data[n],accum[9999]={0},i,j,temp=0;
  for(i=0;i<n;i++)
  {
    temp=i;
    data[i]=x;
    while(temp>=0)
    accum[temp]+=data[temp--];
    for(j=0;j<9999;j++)
    {
      if(accum[j]==0)
        j++;
      else
        {
          int p=accum[j]/10;
          accum[j]=accum[j]%10;
          accum[j+1]+=p;
        }
    }
  }
  int len=9998;
  while(accum[len]==0)
  len--;
  for(i=len;i>=0;i--)
  printf("%d",accum[i]);

}

标签：38,PAT,加强版,temp,int,scanf,d%,printf,accum
来源： https://www.cnblogs.com/yunners/p/12488326.html