how to find soundex number

First, the topic requirements:Given a positive integer in decimal, write down all integers starting with 1, to N, and then count the number of "1" that appears. Requirements: 1. Write a function f (n) and return the number of "1" that appears between 1 and N.　　　 For example F (12) = 5. 2. In the range of 32-bit integers, the maximum n of "f (n) =n" satisfying the condition is what.Second, the solution of id

Problem-solving ideas: To find the number of [a, b] interval and n coprime, we can convert to [1,b] the number of n coprime minus the number of [1,a-1] and N coprimeand the number of n coprime required in [1,b], we can naturally think of the principle of repulsionThe

Find the number of digits of a number factorialFlyfish 2015-8-15For example 7! =5040, the factorial result of 7 is 4 digits (10 binary)Ask for a number of digits1 Loop methodint get_digit_loop(int N) { int0; do { digit ++; while100); return digit;}2 recursive modeint get_digit_recursion(int N){

Freshman competition: Accepted: 15 Submit: 243 Time Limit: 1000 MS Memory limit: 65536 KB Description James had a lot of students who didn't like to think about a + B in the last semi-finals. To apologize, James made a simple A + B Question for everyone, I hope you will be happy to answer questions. So, the question is coming !!!Evaluate the value of X, the smallest positive integer that makes B/(a + x) an integer. Input The first

First, the height of the two-fork treeSimilar to the idea of merging sort. The height of the lowest node is first obtained, and then the height of the higher node is compared respectively. Finally, recursion to the root node to find the height of the root node.Find the height of the binary tree int height () {return getheight (_root);} int getheight (nodeSecond, the number of leaf nodes to findRespectively

The topics are described as follows:To see this topic, the most easily thought of is the violent search method. However, that is not a good way, it is also ACM can not passWay to live.Thought analysis: For a given array, we must first order it small to large. If the minimum number is not1 (greater than 1), the result of the output must be 1. If it is 1, through the current number and the preceding all the

Number of occurrences of 1 in positive numbers from 1 to n Topic: Enter an integer n to find the number of occurrences of 1 in the decimal representation of the n integers from 1 to n. For example, input 12, from 1 to 12 of these integers contains 1 of the number 1, 10, 1 1 and 12, 11 of which have occurred 5 times

Problem: Find the leftmost number of n^n and the rightmost one, N (1Find the right-hand side:Analysis: Actually find the left and right side of a number is quite simple, fast power each time only take the result of the last one to participate in the next operation, take the final result of the lastFind the leftmost one

#includeusing namespacestd;Const intMAXN =500010;intF[MAXN];intA[MAXN];intN;pairint,int>P[MAXN];voidUpdateintXintval) { while(xN) {f[x]+=Val; X+=x-x; }}intQueryintx) { intres=0; while(x>0) {res+=F[x]; X-=x-x; } returnRes;}intMain () { while(SCANF ("%d", n) = =1N) {memset (F,0,sizeof(F)); Memset (A,0,sizeof(a)); for(intI=1; i) {scanf ("%d",P[i].first); P[i].second=i; } sort (P+1, p+n+1); a[p[1].second]=1; for(intI=2; i) { if(p[i].first!=p[i-1].first) a[p[i].second]=

Write a program to find the number of K in the array and the position of the output number. For example, in {2, 4, 3, 7}, the first number is 7 and the position is 4. The numbers of the second and third largest are 4, which can be output at any position of 1 and 3. Function interface: int find_orderk (const int * narry

Given an array of positive integers. All numbers occur even number of times except one number which occurs odd number of times. Find the number in O (n) Time constant space.Example:i/p = [1, 2, 3, 2, 3, 1, 3]o/p = 3A Simple solution are to run nested loops. The outer loop p

/*** Function: Some number of prime factor only 3, 5, 7, find the number of K.*/Two methods:Method One:/** * Idea: Multiply the numbers in the list with 3,5,7 to find the decimal places that are not yet added to the list. * Each time AI is added to the list, a temporary list is used to store 3ai,5ai and 7Ai. To generat

Programmer = programming language basics + data structures + Algorithms Over the past few days, I continue to return to the java basics and learn about the data structure. Here I will implement a simple algorithm-the algorithm for finding the number with the most appearance in the array Public class HashMapTest1 {/*** find the number with the most numbers in an

1, the following is the general method to find the prime number of 1~n: The general method of finding the prime number of 1~n #include 2, for the number of prime screening method, the specific method is not elaborated, Baidu can be found, simply say is to find a prime

Affinity number Time limit:1000ms Memory limit:65536k have questions? Dot here ^_^ The topic describes if the factor of a and equals b,b factor and equals A, and a≠b, is called A, B is the affinity number pair. For example, the sum of all true approximations of 220 (i.e. not their own approximations) is: 1+2+4+5+10+11+20+22+44+55+110=284. 284 of all true approximate and for: 1+2+4+71+142=22

to a group of numbers, only one number appears once, and all the other numbers are paired. Find out this number # include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced."C" gives a group of numbers, only one number appears once, and all the other numbers appea

There is a 3*4 matrix that requires the program to find the value of the largest element in it, along with its line number and column number. (Daleitai algorithm)Solution: Program:#include int main (){int I, j, row = 0, Colum = 0, Max;int A[3][4] = {{1,2,3,4},{4,5,6,7},{-1,3,-5,10}};max = a[0][0];printf ("Array a:\n");for (i = 0; I {for (j = 0; J {printf ("%5d",

Only one number appears once in a set of data. All other numbers appear in pairs. Please find out the number. (using bit operations)Solution: Program:#include int Main (){int arr[] = {1,3,3,1,4,9,4,5,5};int i = 0;int len = sizeof (arr)/ sizeof (Arr[0]); for (i = 1; i {Arr[0] = arr[0] ^ arr[i]; //Put all the numbers in this set of digits to the bitwise OR, becaus