odd精灵圣女+手办
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屈卞18143743920:
数论一题Let m,n∈Z and m odd.Show that (2^m - 1,2^n+1)=1 and (4^m - 1,4^n+1)=1.翻译:m,n是整数,且m是奇数,求证2^m - 1和2^n+1互素.且4^m - 1和4^n+1互... -
27502乐兔
:[答案] 设(2^m-1,2^n+1)=d 所以1=(2^m)^n=(2^n)^m=(-1)^m=-1 (mod d)(此式中等号全为同余符号) 所以d=1或2 又显然d为奇数,所以d=1 得证 后半部分同理可得
屈卞18143743920:
GMAT数学求解If x, y, z areinteger and xy+z is an odd integer, is x an even integer?1) xy+xz is an even integer2) y+xz is an odd integer -
27502乐兔
:[答案] 1)yes 2)not sure因为x为奇数时y,z可一奇数一偶数 偶数时yz全为奇数均是可能情况
屈卞18143743920:
c++数组求平均数 -
27502乐兔
: #include<iostream> using namespace std; #define N 1000 int main(){int tar[N], i;cin>>i;int odd_sum = 0, even_sum = 0;int odd_count = 0, even_count = 0;while(i!=0){tar[even_count+odd_count] = i;if(i %2 == 0){even_sum+=i;even_count+...
屈卞18143743920:
定义一个一维数组,将数组中所有奇数放在另一个数组中,并输出新的奇数数组 -
27502乐兔
: #include#define N 10void move_odd(int a[], int n, int b[], int* odd_num) {int i, j = 0;for (i = 0; i < n; i++) {if ((a[i] & 1) == 1) {b[j] = a[i];j++;}}*odd_num = j; }int main() {int i;int a[N], b[N];int odd_num;for (i = 0; i < N; i++) {scanf(...
屈卞18143743920:
...求下列表达式的值:(1)12+5 div 4 - 13 mod 3 - 2*3(2) succ(15)>15+1(3) ord(true)+ord('5')(4) (5>4) and (7+3f)(6) chr(' ' ' ')(7) odd(4) and odd(15)(8)... -
27502乐兔
:[答案] (1)8 (2)false (3)54 (4)true (5)false (6)' (7)false (8)M succ是求后继,可用于整形、枚举等,比如succ(15)=16,相当于inc(15).ord是求字符的asiil码值.pred求前继,与succ相反.Chr求ascii码的对应字符,这个ASCII码是个1-127的整数.odd判断一个数是否...
屈卞18143743920:
A special sequence is called Harry Potter Sequence.The sequence meets that F[i]+F[i+1]=F[i+2]+F[i+3] for any i>=0.In addition,if an integer is positive and odd,... -
27502乐兔
:[答案] counter没有清0 另外odd是奇数,所以是F[i]%2==1
屈卞18143743920:
用while求1 - 100+奇数和,偶数积+++C语言
27502乐兔
: #include <stdio.h> #include <stdlib.h> //求1到100之间的奇数之和、偶数之积. int main() {int n = 1, sum = 0;double odd = 1;while (n <= 100){if (n % 2 == 1){sum += n;}else{odd *= n;}n++;}printf("sum=%d,odd=%d",sum,odd);system("pause"); } 希望对你有所帮助,
屈卞18143743920:
求解一道Gmat数据充分题If x and y are integers,is y an even integer?(1) 2y – x = x^2 – y^2(2) x is an odd integer.条件(1)单独就是充分的 -
27502乐兔
:[答案] From (1),we can know that x^2+x=y^2+2y.Because x is an odd integer,x^2 is an odd integer too.So the resuit of x^2+2 is an even integer.That is to say y^2+2y is an even integer.If y is an odd integer,y...
屈卞18143743920:
我刚买PS2不久谁能帮我介绍几款中文版的RPG游戏呀?谢了!
27502乐兔
: 最好玩的中文RPG就属“宿命传说2”了,其次“妖精战士 精灵的黄昏”可以玩玩的,其他也没什么中文的RPG了,这些都是港版或台版的,本身中文游戏就很少,更何况RPG,还有个指环王 第3纪元可以试试,但不想半途放弃的话就算了!