Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 15699 Accepted Submission(s): 5715
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?
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.
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 15HintFrom 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.
Author
fatboy_cw@WHU
Source
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题解：把HDU 2089 改改就行了.......我还因为一个if判断条件吃了几发TLE....
AC代码：
//#include<bits/stdc++.h>
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<string.h>
using namespace std;
typedef long long ll;
int dig[63];
ll dp[63][2][63];
ll dfs(int pos,bool have,int last,int flag)
{
int i;
if (pos < 0) return have;
if (flag==0 && dp[pos][have][last]!=-1) return dp[pos][have][last];
int n=flag ? dig[pos] : 9;
ll ans=0;
for(i=0;i<=n;i++)
{
if (last==4 && i==9) ans += dfs(pos-1,true,i,flag && (i==n));
else ans+=dfs(pos-1,have,i,flag && (i==n));
}
if (flag==0)
{
dp[pos][have][last]=ans;
}
return ans;
}
ll solve(ll x)
{
int len = 0;
while(x)
{
dig[len++] = x % 10;
x /= 10;
}
return dfs(len-1,0,0,1);
}
int main()
{
ll x;
int t;
scanf("%d",&t);
while(t--)
{
memset(dp,-1,sizeof(dp));
scanf("%I64d",&x);
printf("%I64d\n",solve(x));
}
return 0;
}