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A program is required to calculate the average of n rational numbers.Input format:Enter the first line to give the positive integer n (≤100), the second line in a1/b1 a2/b2 … the format given n fractional form of the rational number, where the numerator and denominator are all integers in the shaping range, if negative, the negative sign must appear at the front. Output format:The a/b average of n rational numbers is output in the format of a row. Not

-uDisplay in folders in order of storage Search by criteriaLike what:*Instead of one or more charactersOperating mode: Cat/textdir/*.txt?can only replace one characterOperation mode: cat/textdir/*.? Xt~Home directory of the current userOperation mode: LS ~~usernameSpecify a user's home directoryOperation mode: LS ~admin ~+Current working directoryOperation mode: LS ~+~-Previous working directoryOperation mode: LS ~-[0-9]Search for a single digitOperating mode: ls/textdir/854297[0-9][A-z]Search f

Given a batch of integers, each digit of each integer is parsed to find the single digit with the most occurrences. For example, given 3 integers 1234, 2345, 3456, where the maximum number of occurrences is 3 and 4, both occur 3 times.Input format:The input gives a positive integer n (Output format:In a row, press the format "m:n1 n2 ..." Output, where M is the maximum number of times, N1, N2 、...... For th

Given a batch of integers, each digit of each integer is parsed to find the single digit with the most occurrences. For example, given 3 integers 1234, 2345, 3456, where the maximum number of occurrences is 3 and 4, both occur 3 times.Input format:The input gives a positive integer n (Output format:In a row, press the format "m:n1 n2 ..." Output, where M is the maximum number of times, N1, N2 、...... For th

Simple question.#include #include#include#include#include#include#include#includeusing namespacestd;intN;Doublea[100000+Ten];Doubleb[100000+Ten];Doublesum=0;intMain () {scanf ("%d",N); for(intI=1; i1.0* (n-i+1)*i; for(intI=1; i"%LF",A[i]); for(intI=1; iA[i]; printf ("%.2lf\n", sum); return 0;}PAT (Advanced level) 1104. Sum of number segments (20)

) {ans*=Ten; } returnans;}voidPrint (intN) {Switch(n) { Case 0: printf ("Ling"); Break; Case 1: printf ("Yi"); Break; Case 2: printf ("er"); Break; Case 3: printf ("San"); Break; Case 4: printf ("si"); Break; Case 5: printf ("Wu"); Break; Case 6: printf ("Liu"); Break; Case 7: printf ("Qi"); Break; Case 8: printf ("BA"); Break; Case 9: printf ("JIU"); Break; }}intMainvoid) { Charn[101] = {0}; scanf ("%s",N); intsum =0; for(inti =0; I -; i++) { if(N[i]! =' /') {sum+ = N[

LeetCode (20) house robber (maximum number of elements not adjacent to the array), leetcoderobberDescription You are a professional robber planning to rob houses along a street. each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses

/Let the user loop the operation, the user enters a positive integer (0-20), if less than 0 or greater than 20 are prompted to enter an error,If you enter a number between 0-20, then the sum of 0 to this number is printed,Total user Input 3 times, input errors do not count i

2017 Latest (April 20) mobile phone number attribution to database downloadMobile number Segment Database number attribution to the database mobile number segment Unicom segment Signal Segment No. 147th Segment No. 170th (virtual operator)Authoritative Comprehensive and accu

Pat math prime Link: http://pat.zju.edu.cn/contests/pat-b-practise/1007 Let's define DN as: DN = pN + 1-PN, where Pi is the I prime number. Obviously, d1 = 1 and for n> 1, dn is an even number. The "prime number pair conjecture" considers "There are infinite numbers of adjacent prime numbers with a difference of 2 ". Given any positive integer N ( Input Format:

07-0. Write this number (20) time limit of MS memory limit 65536 KB code length limit 8000 B award Program StandardAuthor Chen, YueRead in a natural number n, calculate the sum of their numbers, and write each digit in Hanyu Pinyin.input Format: each test input contains 1 test cases, which gives the value of the natural numbe