HDU 多校联合 6033 6043

http://acm.hdu.edu.cn/showproblem.php?pid=6033

Add More Zero

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 449 Accepted Submission(s): 319


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
The input contains multiple test cases. Each test case in one line contains only one positive integer m, satisfying 1≤m≤105.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

1
64

[align=left]Sample Output[/align]

Case #1: 0 Case #2: 19
10^k>=2^m-1;k=m*(log2/log10) 向下取整

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int main()
{
int m;
int cast=0;
while(scanf("%d",&m)!=EOF)
{
printf("Case #%d: %0.f\n",++cast,floor(m*1.0*log(2.0)/log(10.0)));//floor 向下取整
}
return 0;
}

KazaQ's Socks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 389 Accepted Submission(s): 246


[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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).
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]

3 7
3 6
4 9

[align=left]Sample Output[/align]

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

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
//前n天顺序出现，后来前n-2天顺序出现，n-1，n交替出现
//1.2.3.4.1.2.3.1.2.4.1.2.3.1.2.4........
int main()
{
ll n,m,cast=0,ans;
while(scanf("%lld%lld",&n,&m)!=EOF)
{
printf("Case #%lld: ",++cast);
if(m<=n) printf("%lld\n",m);
else
{
ans=(m-n)/(n-1)%2;
if((m-n)%(n-1)==0) printf("%lld\n",ans==1?n-1:n);
else printf("%lld\n",(m-n)%(n-1));
}
}
return 0;
}