XTU OJ 1210 Happy Number (暴力+打表）

Problem Description

Recently, Mr. Xie learn the concept of happy number. A happy number is a number contain all digit 7 or only 1 digit other than 7. For example, 777 is a happy number because 777 contail all digit 7, 7177 and 87777 both happy number because only 1 digit other than 7. Whereas 887,799 9807,12345, all of them are not happy number. Now Mr. xie want to know for a given integer n, how many number among [1,n] are happy numbers, but counting them one by one is slow, can you help him?

Input

First line an integer t indicate there are t testcases(1≤t≤100). Then t lines follow, each line an integer n(1≤n≤10⁶, n don't have leading zero).

Output

Output case number first, then the answer.

Sample Input

517172030

Sample Output

Case 1: 1Case 2: 7Case 3: 10Case 4: 10Case 5: 11

比赛看到这道题的时候，看到这句话counting them one by one is slow,觉得暴力就可能会超时，当时就没有往那方面去想了，后来没题目做了，就专心搞这道题，用暴力+打表去尝试就过了。比赛的时候还是要敢于尝试啊，只要有想法就要去实现，哪怕想法是错误的，我们也要勇于尝试！只有不断的探索才会有进步；

这道题的思路就是打表，因为数据量还不算太大，打表还是可以做的出；

下面是ac代码；

#include "stdio.h"#include "string.h"const int maxn=1000000+10;int  a[maxn];void init() //打表枚举{    int i;    for(i=0;i<=9;i++) //这里今天做的时候卡在这了，把i的值设为了1，后来检查这个错误查了好久，最后看数据少了几个，才发现这里错了；    {                 //有最后一位为0的情况没有考虑到        a[i]=1;         a[i*10+7]=1; //二位的情况        a[7*10+i]=1;        a[i*100+7*10+7]=1;//三位        a[7*100+i*10+7]=1;        a[7*100+7*10+i]=1;        a[i*1000+7*100+7*10+7]=1;//四位        a[7*1000+i*100+7*10+7]=1;        a[7*1000+7*100+i*10+7]=1;        a[7*1000+7*100+7*10+i]=1;        a[i*10000+7*1000+7*100+7*10+7]=1;        a[7*10000+i*1000+7*100+7*10+7]=1;        a[7*10000+7*1000+i*100+7*10+7]=1;        a[7*10000+7*1000+7*100+i*10+7]=1;        a[7*10000+7*1000+7*100+7*10+i]=1;        a[i*100000+7*10000+7*1000+7*100+7*10+7]=1;        a[7*100000+i*10000+7*1000+7*100+7*10+7]=1;        a[7*100000+7*10000+i*1000+7*100+7*10+7]=1;        a[7*100000+7*10000+7*1000+i*100+7*10+7]=1;        a[7*100000+7*10000+7*1000+7*100+i*10+7]=1;        a[7*100000+7*10000+7*1000+7*100+7*10+i]=1;    }}int main(){    memset(a,0,sizeof(a));    init();    int t,i,count,n,j;    scanf("%d",&t);    for(i=1;i<=t;i++)    {        count=0;       scanf("%d",&n);       for(j=1;j<=n;j++)       {           if(a[j]==1)           count++;       }       printf("Case %d: ",i);       printf("%d\n",count);    }    return 0;}

还有一种写法，感觉内存比上面的要少；

#include<iostream>#include<vector>#include<cstdio>using namespace std;int main(){ //    freopen("a.txt","r",stdin);    int t, n, i, j;    int p[200] = {1,2,3,4,5,6,7,8,9,17,27,37,47,57,67,70,71,72,                  73,74,75,76,77,78,79,87,97,177,277,377,477,577,                  677,707,717,727,737,747,757,767,770,771,772,773,                  774,775,776,777,778,779,787,797,877,977,1777,2777,                  3777,4777,5777,6777,7077,7177,7277,7377,7477,7577,7677,                  7707,7717,7727,7737,7747,7757,7767,7770,7771,7772,7773,7774,                  7775,7776,7777,7778,7779,7787,7797,7877,7977,8777,9777,17777,                  27777,37777,47777,57777,67777,70777,71777,72777,73777,74777,75777,                  76777,77077,77177,77277,77377,77477,77577,77677,77707,77717,77727,                  77737,77747,77757,77767,77770,77771,77772,77773,77774,77775,77776,                  77777,77778,77779,77787,77797,77877,77977,78777,79777,87777,97777,                  177777,277777,377777,477777,577777,677777,707777,717777,727777,737777,                  747777,757777,767777,770777,771777,772777,773777,774777,775777,776777,                  777077,777177,777277,777377,777477,777577,777677,777707,777717,777727,                  777737,777747,777757,777767,777770,777771,777772,777773,777774,777775,                  777776,777777,777778,777779,777787,777797,777877,777977,778777,779777,                  787777,797777,877777,977777};    cin >> t;    for(i = 1; i <= t; ++ i)    {        cin >> n;        for(j = 0; p[j] <= n && j < 189; ++ j);        printf("Case %d: %d\n",i,j);    }    return 0;}