Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 21117    Accepted Submission(s): 7903

 

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?

 

 

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.

 

 

Author

fatboy_cw@WHU

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

 

 

Recommend

zhouzeyong

 

题意:含49 , 套路啊!!!

 

 

 

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
//#define MOD 1000000007
#define bug1 cout <<"bug1"<<endl
#define bug2 cout <<"bug2"<<endl
#define bug3 cout <<"bug3"<<endl
using namespace std;
typedef long long ll;

const int MAX_N=2e5+5;
int bit[100];
ll dp[100][5];

///有49
ll dfs(int pos,int have,bool lead,bool limit){
    if(pos==-1) return (have==2);
    if(!limit && !lead && dp[pos][have]!=-1)   return dp[pos][have];
    int up=limit ? bit[pos] : 9;
    ll ans=0;
    for(int i=0;i<=up;i++){
        int temphave=have;
        if(have==0&&i==4)   temphave=1;
        else if(have==0&&i!=4)  temphave=0;
        else if(have==1 && i==9)    temphave=2;
        else if(have==1 && i==4)    temphave=1;
        else if(have==1 && i!=4)    temphave=0;
        ans+=dfs(pos-1,temphave,i==0&&lead,limit&&bit[pos]==i);
    }
    if(!limit && !lead) dp[pos][have]=ans;
    return ans;
}

ll solve(ll n){
    int pos=0;
    while(n){
        bit[pos++]=n%10;
        n/=10;
    }
    return dfs(pos-1,0,true,true);//最高位肯定有限制
}

int main(void){
    ll n;
    int T;
    cin >>T;
    while(T--){
        scanf("%lld",&n);
        memset(dp,-1,sizeof(dp));
        printf("%lld\n",solve(n));
    }
}


你看!踏马是套路题吧!♪(^∇^*)