Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7926 Accepted Submission(s): 2780
Problem 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?
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
1
50
500
Sample Output
0
1
15
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.
思路：
dp[i][0]表示不含49的个数 。
dp[i][1]表示不含49且高位为9的个数；
dp[i][2]表示包括49的个数；
dp[i][0]=10*dp[i-1][0]-dp[i-1][1]; //不含49的数能够随意加10个数字。减去9前面加4的个数
dp[i][1]=dp[i-1][0]; //不含49的数最高位加9
dp[i][2]=dp[i-1][2]*10+dp[i-1][i]; //包括49的数字能够加入0~9，高位为9的能够加4；
dp[i][1]表示不含49且高位为9的个数；
dp[i][2]表示包括49的个数；
dp[i][0]=10*dp[i-1][0]-dp[i-1][1]; //不含49的数能够随意加10个数字。减去9前面加4的个数
dp[i][1]=dp[i-1][0]; //不含49的数最高位加9
dp[i][2]=dp[i-1][2]*10+dp[i-1][i]; //包括49的数字能够加入0~9，高位为9的能够加4；
#include"stdio.h"
#include"string.h"
#include"iostream"
#include"algorithm"
#include"math.h"
#include"vector"
using namespace std;
#define LL __int64
#define N 25
LL dp[N][3];
void inti()
{
memset(dp,0,sizeof(dp));
dp[0][0]=1;
int i;
for(i=1; i<N; i++)
{
dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
dp[i][1]=dp[i-1][0];
dp[i][2]=dp[i-1][2]*10+dp[i-1][1];
}
}
LL work(LL x)
{
int i,k,flag;
int a[N];
LL ans;
k=0;
while(x)
{
a[++k]=x%10;
x/=10;
}
a[k+1]=ans=flag=0;
for(i=k; i>0; i--)
{
ans+=a[i]*dp[i-1][2];
if(flag)
ans+=a[i]*dp[i-1][0];
else
{
if(a[i]>4)
ans+=dp[i-1][1];
}
if(a[i+1]==4&&a[i]==9)
flag=1;
// printf("%d\n",ans);
}
return ans;
}
int main()
{
int T;
LL n;
inti();
scanf("%d",&T);
while(T--)
{
scanf("%I64d",&n);
printf("%I64d\n",work(n+1));
}
return 0;
}