突然觉得心好累,BestCoder在1500~1700之间一晃就是几个月,但我还是依旧那么水,怎么也跨不过1700的坎

1001 ZYB's Biology

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 0 Accepted Submission(s): 0

Problem Description

After getting

600 scores in

NOIP

ZYB(ZJ?267) begins to work with biological questions.Now he give you a simple biological questions:
he gives you a

DNA sequence and a

RNA sequence,then he asks you whether the

DNA sequence and the

RNA sequence are
matched.

The

DNA sequence is a string consisted of

A,C,G,T ;The

RNA sequence is a string consisted of

A,C,G,U .

DNA sequence and

RNA sequence are matched if and only if

A matches

U ,

T matches

A ,

C matches

G ,

G matches

C on each position.

Input

In the first line there is the testcase

T .

For each teatcase:

In the first line there is one number

N .

In the next line there is a string of length

N ,describe the

DNA sequence.

In the third line there is a string of length

N ,describe the

RNA sequence.

1≤T≤10 ,

1≤N≤100

Output

For each testcase,print

YES or

NO ,describe whether the two arrays are matched.

Sample Input

  
   
    2
4
ACGT
UGCA
4
ACGT
ACGU

Sample Output

  
   
    YES
NO

Source

BestCoder Round #65

/************************************************************************/

附上该题对应的中文题

ZYB's Biology

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 131072/131072 K (Java/Others)

问题描述

 $Z Y B (Z J ? 267)$ 在 $N O I P$ 拿到 $600$ 分之后开始虐生物题,他现在扔给你一道简单的生物题:给出一个 $D N A$ 序列和一个 $R N A$ 序列，
问它们是否配对。 $D N A$ 序列是仅由 $A, C, G, T$ 组成的字符串， $R N A$ 序列是仅由 $A, C, G, U$ 组成的字符串。 $D N A$ 和 $R N A$ 匹配当且仅当每个位置上 $A$ 与 $U$ , $T$ 与 $A$ , $C$ 与 $G$ , $G$ 与 $C$ 匹配。

输入描述

第一行一个整数 $T$ 表示数据组数。对于每组数据：第一行一个整数 $N$ 表示 $D N A$ 和 $R N A$ 序列的长度.第二行一个长度为 $N$ 的字符串 $A$ 表示 $D N A$ 序列.第三行一个长度为 $N$ 的字符串 $B$ 表示 $R N A$ 序列. $\leq T \leq 10$ , $\leq N \leq 100$

输出描述

对于每组数据,输出一行 $Y E S$ 或 $N O$ ,表示是否匹配.

输入样例

2
4
ACGT
UGCA
4
ACGT
ACGU

输出样例

YES
NO

/****************************************************/

出题人的解题思路：

ZYB's Biology

直接按照题意模拟即可.

签到题,if语句判断一下就可以搞定了

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 105;
const int M = 10005;
const int inf = 1000000000;
const int mod = 2009;
char s1[N],s2[N];
int main()
{int t,n,i,j;scanf("%d",&t);while(t--){scanf("%d",&n);scanf("%s",s1);scanf("%s",s2);for(i=j=0;i<n;i++,j++)if(s1[i]=='A'&&s2[j]=='U'||s1[i]=='C'&&s2[j]=='G'||s1[i]=='G'&&s2[j]=='C'||s1[i]=='T'&&s2[j]=='A')continue;elsebreak;if(i!=n)puts("NO");elseputs("YES");}return 0;
}

1002 ZYB's Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 0 Accepted Submission(s): 0

Problem Description

ZYB played a game named

NumberBomb with his classmates in hiking:a host keeps a number in

[1,N] in mind,then
players guess a number in turns,the player who exactly guesses

X loses,or the host will tell all the players that
the number now is bigger or smaller than

X .After that,the range players can guess will decrease.The range is

[1,N] at first,each player should guess in the legal range.

Now if only two players are play the game,and both of two players know the

X ,if two persons all use the best strategy,and the first player guesses first.You are asked to find the number of

X that the second player
will win when

X is in

[1,N] .

Input

In the first line there is the number of testcases

T .

For each teatcase:

the first line there is one number

N .

1≤T≤100000 ,

1≤N≤10000000

Output

For each testcase,print the ans.

Sample Input

1
3

Sample Output

Source

BestCoder Round #65

/************************************************************************/

附上该题对应的中文题

ZYB's Game

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

问题描述

 $Z Y B$ 在远足中,和同学们玩了一个“数字炸弹”游戏：由主持人心里想一个在 $[1, N]$ 中的数字 $X$ ，然后玩家们轮流猜一个数字，如果一个玩家恰好猜中 $X$ 则算负，否则主持人将告诉全场的人当前的数和 $X$ 比是偏大还是偏小，然后猜测的范围就会相应减小，一开始的范围是 $[1, N]$ .每个玩家只能在合法的范围中猜测.现在假设只有两个人在玩这个游戏,并且两个人都已经知道了最后的 $X$ ,若两个人都采取最优策略.求 $\in [1,N]$ 中是后手胜利的 $X$ 数量.

输入描述

第一行一个整数 $T$ 表示数据组数。接下来 $T$ 行，每行一个正整数 $N$ . $\leq T \leq 100000$ , $\leq N \leq 10000000$

输出描述

 $T$ 行每行一个整数表示答案.

输入样例

1
3

输出样例

/****************************************************/

出题人的解题思路：

ZYB's Game

我们会发现这个模型其实是类比于左右两堆石子,每次可以在一堆里取任意多,取完的人胜利.当左右两堆石子相同时,我们可以简单的

构造后手胜利的方法:即在另一堆石子中取走同样多的石子,否则,先手可以取一些石子使得两堆石子相同.所以,当 $N$ 是奇数输出 $1$ ,否则输出 $0$ .

这道题出题人写得还算详细,就算我们一时间想不出来也不要紧,我们完全可以举一下例子,从举例子的过程中发现规律,这是很多时候没有思路的情况下可以做的事情

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 105;
const int M = 10005;
const int inf = 1000000000;
const int mod = 2009;
int main()
{int t,n,i,j;scanf("%d",&t);while(t--){scanf("%d",&n);if(n%2)puts("1");elseputs("0");}return 0;
}

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