补题多校联合训练第一场

Posted 2020-09-30

1001  Add More Zero

 

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 1to 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

1 64

 

Sample Output

Case #1: 0 Case #2: 19

 

思路：注意到不存在 10^k = 2^m10?k??=2?m?? ，所以就是?log?10??2?m???=?mlog?10??2?，这样做的时间复杂度是 O(1) 。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<vector>
 7 #include<set>
 8 #include<string>
 9 #include<sstream>
10 #include<cctype>
11 #include<map> 
12 #include<stack>
13 #include<queue>
14 using namespace std;
15 #define INF 0x3f3f3f3f
16 typedef long long ll;
17 int gcd(int a, int b){return b==0?a:gcd(b,a%b);} 
18 
19 int m; 
20 
21 int main()
22 {
23 //    freopen("input.txt", "r", stdin);
24 //    freopen("output.txt", "w", stdout);    
25     int t = 1;
26     while(scanf("%d", &m) != EOF)
27     {
28         cout << "Case #" << t++ << ": " << (int) (log10(2) * m) << endl;
29     }
30     return 0; 
31 }

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1011  KazaQ‘s Socks

 

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 7 3 6 4 9

 

Sample Output

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

 

思路：打表，找规律即可。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<vector>
 6 #include<set>
 7 #include<string>
 8 #include<sstream>
 9 #include<cctype>
10 #include<map> 
11 using namespace std;
12 #define INF 0x3f3f3f3f
13  
14 int main()
15 {
16 //    freopen("input.txt", "r", stdin);
17 //    freopen("output.txt", "w", stdout);
18     long long n, k;
19     int flag = 0, ans;
20     while(~scanf("%lld%lld", &n, &k))
21     {
22         if(k <= n)    ans = k;
23         else
24         {
25             int x = (k - n) % (2 * (n - 1));
26             if(x < n && x > 0)    ans = x;
27             else if (x == 0)    ans = n;
28             else ans = x - n + 1;
29         }
30         printf("Case #%d: %d\n", ++flag, ans);
31     }
32     return 0;
33 }

View Code