高中学奥赛对高考有用吗?
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我现在上高一,想自学高中数学跟物理的奥赛,但不知道对高考有没有好处,会不会有奥赛类型的高考题啊?
另外,推荐点几本数学奥赛书吧
高考考的题难道都很容易吗?拉分的主要是哪门啊?
一个参加过竞赛的朋友跟我说,高考最难的题目和奥赛的简单题差不多吧。 参考技术B 作用:
1.全国一等奖可以保送(不过去年入学的学生已经享受不到了)
2.其他的奖励可以优先参加自主招生,自主招生考试考竞赛内容为多
3.高考的理科压轴题目,你如果会一些竞赛知识,你会很容易做的 参考技术C 有用是有用的。
1,竞赛获奖可以有保送机会
2,对自主招生有极大的好处
3,退一步讲,即使没有提前招生的机会,对高考难题还是有好处的。 参考技术D 两者几乎无关
高考不会出那么难的题
不过你要是奥赛学好了,高考肯定显得很简单了
高考有难题,但是奥赛题比高考的水平高太多
参考下:我高考理综280分,奥赛几乎没有会做的题
语文最难得高分,其他三门看个人情况
codevs 1277 生活大爆炸 2012年CCC加拿大高中生信息学奥赛
Sheldon and Leonard are physicists who are ?xated on the BIG BANG theory. In order to exchange secret insights they have devised a code that encodes UPPERCASE words by shifting their letters forward.
谢耳朵和莱纳德是研究BB理论的物理学家。要用暗号(大写字母)联系。
Shifting a letter by S positions means to go forward S letters in the alphabet. For example, shifting B by S = 3 positions gives E. However, sometimes this makes us go past Z, the last letter of the alphabet. Whenever this happens we wrap around, treating A as the letter that follows Z. For example, shifting Z by S = 2 positions gives B.
(读懂这段话是解题的关键,翻译了就没意义了)
Sheldon and Leonard’s code depends on a parameter K and also varies depending on the position of each letter in the word. For the letter at position P, they use the shift value of S = 3P + K.
他们有一个密钥K。第P个字母有S = 3P + K。
For example, here is how ZOOM is encoded when K = 3. The ?rst letter Z has a shift value of S = 3 × 1 + 3 = 6; it wraps around and becomes the letter F. The second letter, O, has S = 3 × 2 + 3 = 9 and becomes X. The last two letters become A and B. So Sheldon sends Leonard the secret message: FXAB
Write a program for Leonard that will decode messages sent by Sheldon.
The input will be two lines. The ?rst line will contain the positive integer K (K < 10), which is used to compute the shift value. The second line of input will be the word, which will be a sequence of uppercase characters of length at most 20.
输入有两行。第一行一个正整数K(〈10)。第二行是最多20个大写字母组成的信。
The output will be the decoded word of uppercase letters.
输入就是解密后的信。
样例1:
3
FXAB
样例2:
5
JTUSUKG
样例1:
ZOOM
样例2:
BIGBANG
#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; int main() { char b[25]; int i,k,s; scanf("%d",&k); scanf("%s",b); for(i=0;b[i];i++) { s=3*(i+1)+k; if(b[i]-s<‘A‘) { s-=b[i]-‘A‘; b[i]=‘Z‘+1; } printf("%c",b[i]-s); } }
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