microsoft range else result tdi ask lis positive -1

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：註意到不存在 10^k = 2^m10?k??=2?m?? ，所以就是?log?10??2?m???=?mlog?10??2?，這樣做的時間復雜度是 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，找規律即可。

 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 }

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【補題】多校聯合訓練第一場