2017 Multi-University Training Contest - Team 1(A+K)
2017-07-25 20:30
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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:
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:
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); } }
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