首页 > 代码库 > Counting Game

Counting Game

2024-07-12 21:00:37   223人阅读

There are n people standing in a line, playing a famous game called ``counting". When the game begins, the leftmost person says ``1" loudly, then the second person (people are numbered 1 to n from left to right) says ``2" loudly. This is followed by the 3rd person saying ``3" and the 4th person say ``4", and so on. When the n-th person (i.e. the rightmost person) said ``n" loudly, the next turn goes to his immediate left person (i.e. the (n - 1)-th person), who should say ``n + 1" loudly, then the (n - 2)-th person should say ``n + 2" loudly. After the leftmost person spoke again, the counting goes right again.

There is a catch, though (otherwise, the game would be very boring!): if a person should say a number who is a multiple of 7, or its decimal representation contains the digit 7, he should clap instead! The following tables shows us the counting process for n = 4 (`X‘ represents a clap). When the 3rd person claps for the 4th time, he‘s actually counting 35.

 

Person123432123
Action123456X89
Person432123432
Action10111213X1516X18
Person123432123
Action1920X2223242526X
Person432123432
ActionX293031323334X36

Given n, m and k, your task is to find out, when the m-th person claps for the k-th time, what is the actual number being counted.

 

Input 

There will be at most 10 test cases in the input. Each test case contains three integers n, m and k ( 2$ \le$n$ \le$100, 1$ \le$m$ \le$n, 1$ \le$k$ \le$100) in a single line. The last test case is followed by a line with n = m = k = 0, which should not be processed.

 

Output 

For each line, print the actual number being counted, when the m-th person claps for the k-th time. If this can never happen, print ` -1‘.

 

Sample Input 

 

4 3 14 3 24 3 34 3 40 0 0

 

Sample Output 

 

17212735



分析：报数问题，凡是7的倍数或者含7的数就不喊出来。n个人，当第m个人喊这类数p次后输出他们一共喊了多少次。
     n+n-2代表循环报数是一个循环有这么多次  如12345432 少了一个5和一个1  换为n也是如此比较

代码:
#include<stdio.h>#include<string.h>#define N 1000001int pu(int x){    if(x%7==0)      return 1;    while(x)    {        if(x%10==7)          return 1;        x/=10;    }    return 0;}int main(){    int n,m,k,i,j,sum,count[10001];    while(~scanf("%d %d %d",&n,&m,&k))    {        sum=0;        if(n==0&&m==0&&k==0)           break;        for(i=1;i<=n;i++)           count[i]=i;        j=n-1;        for(i=n+1;i<n+n-2;i++)        {            count[i]=j;            j--;        }        for(i=1;i<N;i++)        {            if(pu(i)==1)            {                if(i%(n+n-2)==0)                {                    if(count[n+n-2]==m)                       sum++;                 }                else                {                    if(count[i%(n+n-2)]==m)                       sum++;                }            }            if(sum==k)            {                printf("%d\n",i);                break;            }        }        if(i==N)           printf("-1\n");    }    return 0;}


可参考的代码：

#include<set>#include<map>#include<stack>#include<queue>#include<ctime>#include<cmath>#include<vector>#include<string>#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int n, m, t;bool seven(int k){    if (!(k % 7))return 1;    while (k){        if ((k % 10) == 7)return 1;        k /= 10;    }    return 0;}void solve(){    int x = m;    int cnt=0;    int step = 0;    int k1 = 2 * (n - m);//如果向右走回到自己 步数是2*(n-m)    int k2 = 2 * (m - 1);//向左走回到自己是2*（m-1)    if (seven(x))cnt++;    if (t == cnt){ printf("%d\n", x); return;}    while (x <= 100000000){        step++;        if (step % 2 && k1 == 0)continue;        if (!(step % 2) && k2 == 0)continue;        if (step % 2)x += k1;        else x += k2;        if (seven(x))cnt++;        if (cnt == t){ printf("%d\n", x); return; }    }}int main(){    while (1){        scanf("%d%d%d", &n, &m, &t);        if (!n)break;        solve();    }    return 0;}


联系
我们