Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15102 Accepted Submission(s): 5452
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.
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.
Author
fatboy_cw@WHU
Source
/*
这个是自己的代码反着求的，正如“不要49”，跑了65ms
*/
#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<string.h>
#include<cstdio>
#define N 22
using namespace std;
long long t,n;
int g[N];//用来存放数字的位数
long long dp[N][N];//dp[i][j][k]表示当前剩余位数是i上一位数字是j且当前位上是否为4的状态的个数
long long dfs(int len,bool s,bool f)//s表示是不是当前位上是不是4
{
if(len<)
return ;
if(!f&&dp[len][s]!=-)
return dp[len][s];
int fmax=f?g[len]:;
long long cur=;
//cout<<"fmax="<<fmax<<endl;
for(int i=;i<=fmax;i++)
{
if(s&&i==) continue;//不要有49
cur+=dfs(len-,i==,f&&(i==fmax));
//cout<<"cur="<<cur<<endl;
}
//cout<<cur<<endl;
if(!f)
dp[len][s]=cur;
return cur;
}
long long solve(long long x)
{
int len=;
while(x!=)
{
g[len++]=x%;
x/=;
}
//for(int i=1;i<=len;i++)
// cout<<g[i];
//cout<<endl;
memset(dp,-,sizeof dp);
return dfs(len-,false,true);
}
int main()
{
//freopen("in.txt","r",stdin);
scanf("%lld",&t);
while(t--)
{
scanf("%I64d",&n);
printf("%I64d\n",n-solve(n)+);
}
return ;
}
/*
一开始真的按照“不要49”这么来求的。写博客的时候看看别人博客吸收精华，下面是正着求得，z[i]数组真是神来之笔
本来我也行正着求但是不知道怎么表示找到49之后应该加什么，一个z[i]完美解决了找到49之后应该加多少
*/
#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<string.h>
#include<cstdio>
#define N 30
using namespace std;
long long t,n;
int g[N];//用来存放数字的位数
long long dp[N][];//dp[i][j][k]表示当前剩余位数是i上一位数字是j且当前位上是否为4的状态的个数
long long z[N]={};
long long dfs(int len,bool s,bool f)//s表示是不是当前位上是不是4
{
if(len==)
return ;
if(!f&&dp[len][s]>=)
return dp[len][s];
int fmax=f?g[len]:;
long long cur=;
//cout<<"fmax="<<fmax<<endl;
for(int i=;i<=fmax;i++)
{
if(s&&i==)
{
cur+=f?n%z[len-]+:z[len-];//当前位找到了，剩下的不用搜了，直接加上就行了，这里加的是剩下的所有位
}
else
cur+=dfs(len-,i==,f&&g[len]==i);
//cout<<"cur="<<cur<<endl;
}
//cout<<cur<<endl;
return f?cur:dp[len][s]=cur;
}
long long solve(long long x)
{
int len=;
while(x)
{
g[++len]=x%;
x/=;
}
//for(int i=1;i<=len;i++)
// cout<<g[i];
//cout<<endl;
g[len+]=;
return dfs(len,false,true);
}
int main()
{
//freopen("in.txt","r",stdin);
for (int i=;i<N;i++)
{
z[i]=z[i-]*;
}
scanf("%lld",&t);
memset(dp,-,sizeof dp);
while(t--)
{
scanf("%lld",&n);
printf("%lld\n",solve(n));
}
return ;
}