HDU 3555 Bomb [digit]

HDU 3555 Bomb [digit]

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 between des the sub-sequence "49", the power of the blast wocould 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

 3150500 

Sample Output

 0115 

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.         

 

 

Analyse:

Same as "not 62", digital template !!

However, long is not enabled in the brain;

Code:

 

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            #define INF 0x7fffffff#define SUP 0x80000000#define mem(a,b) memset(a,b,sizeof(a))using namespace std;typedef __int64 LL;const int N=100007;LL dp[30][10][2];int digit[30];LL dfs(int pos,int pre,int en,int limit){ if(pos==-1) return en; if(!limit&&dp[pos][pre][en]!=-1) return dp[pos][pre][en]; int last=limit?digit[pos]:9; LL ret=0; for(int i=0;i<=last;i++) { ret+=dfs(pos-1,i,en||(pre==4&&i==9),limit&&i==last); } if(!limit) dp[pos][pre][en]=ret; return ret;}LL solve(LL x){ int cnt=0; while(x) { digit[cnt++]=x%10; x/=10; } return dfs(cnt-1,0,0,1);}int main(){ int T;scanf("%d",&T); LL n; mem(dp,-1); while(T--) { scanf("%I64d",&n); printf("%I64d\n",solve(n)); } return 0;}