Description
The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the
power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?
Input
The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.
The input terminates by end of file marker.
Output
For each test case, output an integer indicating the final points of the power.
Sample Input
3 1 50 500
Sample Output
0 1 15
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.
题意：求1到n一共有多少个49
一个简单的数位DP题。
<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;
}