HDU3555:Bomb

题目描述

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?

输入格式

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.

输出格式

For each test case, output an integer indicating the final points of the power.

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翻译：给出n，问1~n中有多少个数中含有49.

A掉的第一道数位dp题。为了方便(其实是另外一个OJ的翻译)，我们把带有49的数叫做幸运数。设dp(i,0)表示长度为i的幸运数个数，其中这些数不以49开头。dp(i,1)就是以49开头的情况。然后记搜即可。具体细节在代码里标明：

#include<iostream>
#include<cstring>
#include<cstdio>
#define maxn 21
using namespace std;
 
int N[maxn];
long long dp[maxn][2];
 
long long dfs(int len,bool is4,bool havelim){//havelim : have limitation
    if(len==0) return 1;//搜索范围为1~tmp
    if(!havelim&&dp[len][is4]) return dp[len][is4];//只有没有限制时才可以记忆化。为什么？因为记的就是没有限制的情况......
    int lim=havelim?N[len]:9;//如果上一位填满了限制，那么这一位也必须有限制
    long long cnt=0;//有多少个不是幸运数的数
    for(register int i=0;i<=lim;i++){
        if(is4&&i==9) continue;//排除幸运数的情况
        cnt+=dfs(len-1,i==4,havelim&&i==lim);//只有上一位有限制并且这一位达到限制时下一位才能有限制
    }
    return havelim?cnt:dp[len][is4]=cnt;//没有限制时记忆化
}
 
int main(){
    int t;cin>>t;
    while(t--){
        memset(N,0,sizeof N);
        memset(dp,0,sizeof dp);
        unsigned long long n,tmp; cin>>n; tmp=n; int k=0;
        while(tmp) N[++k]=tmp%10,tmp/=10;//得到n的位数
        printf("%lld\n",n-dfs(k,false,true)+1);//得到“非幸运数”个数应该比得到“幸运数”个数要容易实现
    }
    return 0;
}