POJ3208:Apocalypse Someday

题目描述

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.

输入格式

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.

输出格式

For each test case, your program should output the nth beastly number.

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翻译：给出n，问1~n中含有666的数的个数

很明显的数位动规。我们用记忆化搜索来做，先设计状态。

len,pre1,pre2表示当前为第len位，前一位是否为6，前第二位是否为6。为了减少思维复杂度，我们可以数位动规算出有多少不含666的数的个数。那么设(len,pre1,pre2)表示不含666的数的个数，可以很容易写出代码：

long long dfs(int len,bool pre1,bool pre2,bool havelim){
    if(len==0) return 1;
    if(!havelim&&dp[len][pre1][pre2]) return dp[len][pre1][pre2];
    int lim=(havelim?bit[len]:9);
    long long cnt=0;
    for(register int i=0;i<=lim;i++){
        if(pre1&&pre2&&i==6) continue;
        cnt+=dfs(len-1,(i==6),pre1,(havelim&&i==lim));
    }
    return havelim?cnt:dp[len][pre1][pre2]=cnt;
}

完整代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

long long dp[21][2][2];
int n,bit[21];

inline int read(){
    register int x(0),f(1); register char c(getchar());
    while(c<'0'||'9'<c){ if(c=='-') f=-1; c=getchar(); }
    while('0'<=c&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}

long long dfs(int len,bool pre1,bool pre2,bool havelim){
    if(len==0) return 1;
    if(!havelim&&dp[len][pre1][pre2]) return dp[len][pre1][pre2];
    int lim=(havelim?bit[len]:9);
    long long cnt=0;
    for(register int i=0;i<=lim;i++){
        if(pre1&&pre2&&i==6) continue;
        cnt+=dfs(len-1,(i==6),pre1,(havelim&&i==lim));
    }
    return havelim?cnt:dp[len][pre1][pre2]=cnt;
}

inline bool check(long long x){
    int d=0; long long tmp=x;
    while(x) bit[++d]=x%10,x/=10;
    return (tmp+1-dfs(d,false,false,true))>=n;
}

int main(){
    int t=read();
    while(t--){
        n=read();
        long long l=0,r=1e18,mid=0,ans=0;
        while(l<=r){
            mid=l+r>>1;
            if(check(mid)) ans=mid,r=mid-1;
            else l=mid+1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}