Description
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?
Input
The input terminates by end of file marker.
Output
Sample Input
Sample Output
Hint
From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499", so the answer is 15.
<pre name="code" class="plain">#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
__int64 dp[22][4],n;
int b[22];
//dp[i][0]表示长度为i,包括49的个数
//dp[i][1]表示长度为i,开头为9的个数
//dp[i][2]表示长度为i,没有49
void fun()
{
int i;
memset(dp,0,sizeof(dp));
dp[0][2]=1;
for(i=1;i<20;i++)
{
dp[i][0]=(__int64)dp[i-1][0]*10+dp[i-1][1];//有49=上一位有9的 + 上一位有 49 *10
dp[i][1]=dp[i-1][2];//有 9的 =上一位无 49的
dp[i][2]=(__int64)dp[i-1][2]*10-dp[i-1][1];// 没有49的 =上一位无 49*10-上一位有9的
}
}
int main()
{
int t,i;
fun();
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
scanf("%I64d",&n);
n++;
int len=0;
while(n)
{
b[++len]=n%10;
n/=10;
}
b[len+1]=0;
bool f=0;
__int64 ans=0;
for(i=len;i>0;i--)
{
ans+=(__int64)b[i]*dp[i-1][0];
if(f) ans+=(__int64)dp[i-1][2]*b[i];
if(!f&&b[i]>4) ans+=dp[i-1][1];
if(b[i]==9&&b[i+1]==4) f=1;
}
printf("%I64d\n",ans);
}
}
return 0;
}