A:A题地址
题意:对于给定的n,找到最大的k使得10^k<=2^n.
题解:等式两边同时取log(10)变成k<=log(10)2^n.
代码:
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<vector>
#include<queue>
#include<algorithm>
#include<map>
#include<math.h>
#define NI freopen("in.txt","r",stdin);
#define NO freopen("out.txt","w",stdout);
using namespace std;
typedef long long int ll;
typedef pair<int,int>pa;
const int N=1e5+10;
const int MOD=1e9+7;
const ll INF=1e18;
int read()
{
int x=0;
char ch = getchar();
while('0'>ch||ch>'9')ch=getchar();
while('0'<=ch&&ch<='9')
{
x=(x<<3)+(x<<1)+ch-'0';
ch=getchar();
}
return x;
}
/************************************************************/
int n;
int main()
{
int cas=1;
while(cin>>n)
{
int ans=(int)(log10(2)*n);
printf("Case #%d: %d\n",cas++,ans);
}
}
K:K题地址
题意:一个人有编号1-n双袜子。每天早上,他从袜子堆中选出编号最小的穿,晚上就丢在洗衣机中,当洗衣机内的袜子数==n-1时,洗衣机工作,明天晚上袜子重新放到袜子堆中,问第k天他会穿哪双袜子?
题解:
随便写一下就会看出规律。例如当n==5时
1234
5123
4123
5123
4123
5123
4123。
注意n==2.
代码:
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<vector>
#include<queue>
#include<algorithm>
#include<map>
#define NI freopen("in.txt","r",stdin);
#define NO freopen("out.txt","w",stdout);
using namespace std;
typedef long long int ll;
typedef pair<int,int>pa;
const int N=1e5+10;
const int MOD=1e9+7;
const ll INF=1e18;
int read()
{
int x=0;
char ch = getchar();
while('0'>ch||ch>'9')ch=getchar();
while('0'<=ch&&ch<='9')
{
x=(x<<3)+(x<<1)+ch-'0';
ch=getchar();
}
return x;
}
/************************************************************/
ll n,k;
int main()
{
int cas=1;
while(cin>>n>>k)
{
ll ans;
if(k<=n-1) ans=k;
else if(n==2) ans=k%2==1?1:2;
else
{
k-=(n-1);
ans=k%(n-1);
if(ans==1)
{
if((k/(n-1))%2==1)
ans=n-1;
else
ans=n;
}
else if(ans==0)
ans=n-2;
else
ans--;
}
printf("Case #%d: %I64d\n",cas++,ans);
}
}