ZOJ Problem Set - 2965

 Accurately Say "CocaCola"!

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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In a party held by CocaCola company, several students stand in a circle and play a game.

One of them is selected as the first, and should say the number 1. Then they continue to count number from 1 one by one (clockwise). The game is interesting in that, once someone counts a number which is a multiple of 7 (e.g. 7, 14, 28, ...) or contains the digit '7' (e.g. 7, 17, 27, ...), he shall say "CocaCola" instead of the number itself.

For example, 4 students play this game. At some time, the first one says 25, then the second should say 26. The third should say "CocaCola" because 27 contains the digit '7'. The fourth one should say "CocaCola" too, because 28 is a multiple of 7. Then the first one says 29, and the game goes on. When someone makes a mistake, the game ends.

During a game, you may hear a consecutive of p "CocaCola"s. So what is the minimum number that can make this situation happen?

For example p = 2, that means there are a consecutive of 2 "CocaCola"s. This situation happens in 27-28 as stated above. 27 is then the minimum number to make this situation happen.

Input

Standard input will contain multiple test cases. The first line of the input is a single integerT (1 <= T <= 100) which is the number of test cases. And it will be followed byT consecutive test cases.

There is only one line for each case. The line contains only one integer p (1 <=p <= 99).

Output

Results should be directed to standard output. The output of each test case should be a single integer in one line, which is the minimum possible number for the first of thep "CocaCola"s stands for.

Sample Input

223

Sample Output

2770

题意：就是一个数7的游戏，统计含有7或者是7的倍数的连续数有n个的数列中最小的那个数
思路：按照题意打表即可
#include <stdio.h>  #include <string.h>  #include <algorithm>  using namespace std;    int a[800],len = 0;    int check(int n)  {      if(n%7 == 0)      return 1;      while(n)      {          int r = n%10;          n/=10;          if(r == 7)          return 1;      }      return 0;  }    int main()  {      int t,n,ans,i,j;       for(i = 1;i<=800;i++)      {          if(check(i))          a[i] = 1;          else          a[i] = 0;      }      scanf("%d",&t);      while(t--)      {          scanf("%d",&n);          for(i = 7;i<=800;i++)          {              for(j = i;j<i+n;j++)  //连续的             {                  if(!a[j])                  break;              }              if(j>=i+n)              {                  ans = i;                  break;              }          }          printf("%d\n",ans);      }        return 0;  }