补题： 1001. Add More Zero 1011KazaQ's Socks

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先补比较简单的1001 和1011，最后再来看1002

1001

Add More Zero

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2245 Accepted Submission(s): 1053

Problem Description

There is a youngster known for amateur propositions concerning several mathematical hard problems.

Nowadays, he is preparing a thought-provoking problem on a specific type of supercomputer which has ability to support calculations of integers between

0 and

(2m−1) (inclusive).

As a young man born with ten fingers, he loves the powers of

10 so much, which results in his eccentricity that he always ranges integers he would like to use from

1 to

10k (inclusive).

For the sake of processing, all integers he would use possibly in this interesting problem ought to be as computable as this supercomputer could.

Given the positive integer

m, your task is to determine maximum possible integer

k that is suitable for the specific supercomputer.

Input

The input contains multiple test cases. Each test case in one line contains only one positive integer

m, satisfying

1≤m≤105.

Output

For each test case, output "Case #x:y" in one line (without quotes), where

x indicates the case number starting from

1 and

y denotes the answer of corresponding case.

Sample Input

Sample Output

Case #1: 0Case #2: 19

题目分析：

$答案就是：$ ，注意到不存在 $10^k = 2^m$ ，所以

#include<iostream>#include<cstdio>using namespace std;int main(){    double a=0.301029995663;    int m;    int l=0;    while(scanf("%d",&m)!=EOF)    {        l++;        cout<<"Case #"<<l<<": "<<(int)(m*a)<<endl;    }    return 0;}

KazaQ's Socks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2375 Accepted Submission(s): 1074

Problem Description

KazaQ wears socks everyday.

At the beginning, he has

n pairs of socks numbered from

1 to

n in his closets.

Every morning, he puts on a pair of socks which has the smallest number in the closets.

Every evening, he puts this pair of socks in the basket. If there are

n−1 pairs of socks in the basket now, lazy KazaQ has to wash them. These socks will be put in the closets again in tomorrow evening.

KazaQ would like to know which pair of socks he should wear on the

k-th day.

Input

The input consists of multiple test cases. (about

2000)
For each case, there is a line contains two numbers

n,k

(2≤n≤109,1≤k≤1018).

Output

For each test case, output "Case #x:y" in one line (without quotes), where

x indicates the case number starting from

1 and

y denotes the answer of corresponding case.

Sample Input

3 73 64 9

Sample Output

Case #1: 3Case #2: 1Case #3: 2

题意：KazaQ 有n双袜子，按1到n编号放在柜子里，他早上选一双穿，号码小的优先，晚上把穿的袜子放篮子里，当篮子里的袜子达到n-1双时，他会洗袜子，第二天晚上会把洗过的袜子放到柜子里，问第k天他穿的是哪双袜子？

题目分析：

n=3时

n k ans

3 1 1

3 2 2

3 3 3

3 4 1

3 5 2

3 6 1

3 7 3

3 8 1

3 9 2

3 10 1

3 11 3

。。。。。。

n=4时

n k ans

4 1 1

4 2 2

4 3 3

3 4 4

4 5 1

4 6 2

4 7 3

4 8 1

4 9 2

4 10 4

4 11 1

4 12 2

4 13 3

4 14 1

4 15 2

4 16 4

观察规律：前n天，第i天袜子是i,之后,天数-n,即i-n, n-1为周期，第i天， (i-n)%(n-1) 不为0，袜子为(i-n)%(n-1)，否则 (i-n)/(n-1)为奇数袜子n-1 ；

(i-n)/(n-1)为偶数，袜子为n。

AC代码

#include<cstdio>#include<iostream>using namespace std;int main(){    long long n,k;    int t=0;    while(scanf("%lld%lld",&n,&k)!=EOF)    {        t++;        cout<<"Case #"<<t<<": ";        if(k<=n)        {            cout<<k<<endl;            continue;        }        else        {            long long m=(k-n)%(n-1);            if(m==0)                m=n;            long long rare=(k-n)/(n-1);            if(m<n-1)            {                cout<<m<<endl;            }            else            {                if(rare%2)                    cout<<n-1<<endl;                else                    cout<<n<<endl;            }        }    }    return 0;}

$\left \lfloor \log_{10}(2^m - 1) \right \rfloor = \left \lfloor \log_{10}{2^m} \right \rfloor = \left \lfloor m \log_{10}{2} \right \rfloor$

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