Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 23059    Accepted Submission(s): 8673

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

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.

 

Author

fatboy_cw@WHU

 

Source

 

代码：

#include
      
       #include
       
        #include
        
         using namespace std;typedef long long ll;typedef unsigned long long ull;#define mod 1000000007ll n,m;ll dp[25][3];int bit[25];void init(){    //dp初始值    dp[0][0]=1;    dp[0][1]=dp[0][2]=0;    for(int i=1;i<=20;i++)    {        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];//除去49        dp[i][1]=dp[i-1][0];//首位为9        dp[i][2]=dp[i-1][2]*10+dp[i-1][1];//    }}ll solve(ll a){    int len=0;    while(a!=0)    {        bit[++len]=a%10;        a/=10;    }    bit[len+1]=0;//防止越位    ll ans=0;    bool flag=false;    for(int i=len;i>=1;i--)    {        ans+=dp[i-1][2]*bit[i];        if(flag) ans+=dp[i-1][0]*bit[i];        if(!flag&&bit[i]>4) ans+=dp[i-1][1];        if(bit[i+1]==4&&bit[i]==9)flag=true;    }    if(flag)ans++;//加上a    return ans;}int main(){    ios_base::sync_with_stdio(0);    cin.tie(0);    init();    int T;    cin>>T;    while(T--)    {        cin>>n;        cout<
         
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