poj 3208 Apocalypse Someday 数位dp+二分答案

Apocalypse Someday Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 2203   Accepted: 1110 Description The number 666 is considered to be the occult “number of the beast” and is a well used number in all major apocalypse themed blockbuster movies. However the number 666 can’t always be used in the script so numbers such as 1666 are used instead. Let us call the numbers containing at least three contiguous sixes beastly numbers. The first few beastly numbers are 666, 1666, 2666, 3666, 4666, 5666… Given a 1-based index n, your program should return the nth beastly number. Input The first line contains the number of test cases T (T ≤ 1,000). Each of the following T lines contains an integer n (1 ≤ n ≤ 50,000,000) as a test case. Output For each test case, your program should output the nth beastly number. Sample Input 3 2 3 187 Sample Output 1666 2666 66666 Source POJ Monthly--2007.03.04, Ikki, adapted from TCHS SRM 2 ApocalypseSomeday

 

题意：

求第nn小的数位中含有三个连续的6的数。

简单数位DP+二分

#include
#include
#include

using namespace std;
typedef long long ll;
ll dp[30][30];
int a[30];
ll dfs(int pos,int sta,bool limit)
//计算
{
	if(pos==-1){
		return sta==3;
	}
	if(!limit&&dp[pos][sta]!=-1) 
	return dp[pos][sta];
	int up=limit?a[pos]:9;
	ll tmp=0;
	for(int i=0;i<=up;i++)
	{
	 if(sta==3)
	 tmp+=dfs(pos-1,sta,limit&&i==up);
	 else if(i==6)
	 tmp+=dfs(pos-1,sta+1,limit&&i==up);
	 else
	 tmp+=dfs(pos-1,0,limit&&i==up);
	}
	if(!limit) 
	dp[pos][sta]=tmp;
	return tmp; 
}
 
ll solve(ll x)
{
	ll t=x;
    int pos=0;
    while(x)
    {
        a[pos++]=x%10;
        x/=10;
    }
    return dfs(pos-1,0,true);
    
}
int main()
{
	
    memset(dp,-1,sizeof(dp));
    int T;
    cin>>T;
    for(int i=1;i<=T;i++)
    {
	ll n,m;
	scanf("%lld",&n);
	ll l=666,r=1e18,mid;
	while(l<=r)
	{
	
	 mid=(l+r)/2;
	 if(solve(mid)>=n)
	 	r=mid-1;
	 else
	 	l=mid+1;
	 	
	}
	printf("%lld\n",l);
    }

    return 0;
}