數(shù)字運(yùn)算，判斷一個(gè)數(shù)是否接近素?cái)?shù)
A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in base b is a Niven number if the sum of its digits divides its value.
Given b (2 <= b <= 10) and a number in base b, determine whether it is a Niven number or not.
Input
Each line of input contains the base b, followed by a string of digits representing a positive integer in that base. There are no leading zeroes. The input is terminated by a line consisting of 0 alone.
Output
For each case, print "yes" on a line if the given number is a Niven number, and "no" otherwise.
Sample Input
10 111
2 110
10 123
6 1000
8 2314
0
Sample Output
yes
yes
no
yes
no
標(biāo)簽：
數(shù)字
運(yùn)算
上傳時(shí)間：
2015-05-21
上傳用戶：daguda
編程題（15_01.c）
結(jié)構(gòu)
struct student
{
long num
char name[20]
int score
struct student *next
}
鏈表練習(xí)：
(1).編寫函數(shù)struct student * creat(int n),創(chuàng)建一個(gè)按學(xué)號(hào)升序排列的新鏈表,每個(gè)鏈表中的結(jié)點(diǎn)中
的學(xué)號(hào)、成績(jī)由鍵盤輸入，一共n個(gè)節(jié)點(diǎn)。
(2).編寫函數(shù)void print(struct student *head),輸出鏈表，格式每行一個(gè)結(jié)點(diǎn)，包括學(xué)號(hào)，姓名,分?jǐn)?shù)。
(3).編寫函數(shù)struct student * merge(struct student *a,struct student *b), 將已知的a,b兩個(gè)鏈表
按學(xué)號(hào)升序合并，若學(xué)號(hào)相同則保留成績(jī)高的結(jié)點(diǎn)。
(4).編寫函數(shù)struct student * del(struct student *a,struct student *b),從a鏈表中刪除b鏈表中有
相同學(xué)號(hào)的那些結(jié)點(diǎn)。
(5).編寫main函數(shù)，調(diào)用函數(shù)creat建立2個(gè)鏈表a,b,用print輸出倆個(gè)鏈表；調(diào)用函數(shù)merge升序合并2個(gè)
鏈表，并輸出結(jié)果；調(diào)用函數(shù)del實(shí)現(xiàn)a-b,并輸出結(jié)果。
a:
20304,xxxx,75,
20311,yyyy,89
20303,zzzz,62
20307,aaaa,87
20320,bbbb,79
b：
20302,dddd,65
20301,cccc,99
20311,yyyy,87
20323,kkkk,88
20307,aaaa,92
20322,pppp,83
標(biāo)簽：
student
struct
score
long
上傳時(shí)間：
2016-04-13
上傳用戶：zxc23456789
