pat 1049 Counting Ones

第4，6个点超时。后来发现居然是编程之美的一道原题，用了书上代码，直接AC。

代码：

 

//1049 20:14-20:35
#include<stdio.h>
const int NUM=100000000;
int find(int x)
{
int res=0;
while(x){
if(x%10==1)
res++;
x=x/10;
}
return res;
}

int find1(int x)
{

}
int a[NUM];

int main()
{
int i,j,n;
scanf("%d",&n);
a[0]=0;
for(i=1;i<=n;i++){
a[i]=a[i-1]+find(i);

}
printf("%d",a
);
return 0;
}

AC代码：

//1049 22:31
#include<stdio.h>
int main()
{
int n;
int iCount=0;
int iFactor=1;
int iLowerNum=0;
int iCurrNum=0;
int iHigherNum=0;
scanf("%d",&n);
while(n/iFactor!=0)
{
iLowerNum = n -(n/iFactor) *iFactor;
iCurrNum = (n/iFactor)%10;
iHigherNum = n/(iFactor *10);

switch(iCurrNum){
case 0:
iCount+=iHigherNum * iFactor;
break;
case 1:
iCount +=iHigherNum*iFactor + iLowerNum +1;
break;
default:
iCount +=(iHigherNum+1) *iFactor;
break;
}
iFactor*=10;
}
printf("%d",iCount);
return 0;
}