Algorithm 具有不同相邻数字的最小较大数字
我们是一个给定的数字Algorithm 具有不同相邻数字的最小较大数字,algorithm,data-structures,number-theory,Algorithm,Data Structures,Number Theory,我们是一个给定的数字n1我的解决方案是基于相邻的数字,当我们遇到“99”时有额外的逻辑。我真的不太喜欢它,但也许它会解决你的问题,或者帮助你写出更好的解决方案 public static int Test(string nStr) { var n = int.Parse(nStr); var needRestartCheck = true; while (needRestartCheck) { needRestartCheck = false;
n
1我的解决方案是基于相邻的数字,当我们遇到“99”时有额外的逻辑。我真的不太喜欢它,但也许它会解决你的问题,或者帮助你写出更好的解决方案
public static int Test(string nStr)
{
var n = int.Parse(nStr);
var needRestartCheck = true;
while (needRestartCheck)
{
needRestartCheck = false;
for (var i = 0; i < nStr.Length - 1; i++)
{
var first = nStr[i];
var second = nStr[i + 1];
if (first == second)
{
n += (int)Math.Pow(10, nStr.Length - 2 - i);
nStr = n.ToString();
needRestartCheck |= first == '9';
}
}
}
return n;
}
公共静态int测试(字符串nStr)
{
var n=int.Parse(nStr);
var needRestartCheck=true;
while(needRestartCheck)
{
needRestartCheck=false;
对于(变量i=0;i
我的方法如下,它只适用于我自己解决的问题
#include <bits/stdc++.h>
#define ll unsigned long long
using namespace std;
bool noDec(string str,int i,ll num){
str[i+1]--;
return stoll(str)>num;
}
ll num;
string str;
int main() {
cin>>num;
str=to_string(num+1);
for(int i=0; i<str.length()-1; i++){
if(str[i]==str[i+1]){
if(str[i]!='9'){
if(noDec(str,i,num)&&str[i]!='0')str[i+1]--;
else str[i+1]++;
}
else{
if(i==0){
str='1'+string(str.length(),'0');
}else{
str[i-1]++;
for(int j=i; j<str.length(); j++){
str[j]='0';
}
}
}i=0;
}
}
if(str[0]==str[1]&&str[0]=='9'){
str=string((ll)log10(num)+2,'1');
for(int i=1; i<str.length(); i+=2)
str[i]='0';
}
cout<<str;
}
#包括
#定义ll unsigned long long
使用名称空间std;
bool noDec(字符串str、int i、ll num){
str[i+1]-;
返回stoll(str)>num;
}
llnum;
字符串str;
int main(){
cin>>num;
str=到_字符串(num+1);
对于(int i=0;这对于大数字来说速度较慢的原因是,如果您当前正在尝试100000000,那么在最左边的两个零不再相同之前,您将添加大量的1。尝试搜索最左边的一对相等的数字并直接更改它们。但是如果给定的数字是99999999,那么更改相邻的数字将如何uld工作?您可以通过在正确的位置添加1来更改它们,这样它将勾选到100999999。然后清除右边的数字以获得100000000,然后继续;最左边的相等对是00,所以在正确的位置添加1以获得101000000,依此类推。谢谢,但这篇文章是关于slowness@SandroJologua,我已经用alt解决了速度问题ernate解决方案。当你给这个问题[数论]加上标签,然后把它变成一个字符串操作问题时,我喊“犯规”!它也可以很容易地变成字母。这可以用数学来解决。在我对你的原始代码的更正中,我添加了第二个例子,它的速度和这个一样快。
bool valid(int x) {
unsigned long long n;
if (valid(n)) {
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
bool valid(unsigned long long x) {
unsigned long long remainder = x % 10;
while (x) {
unsigned long long quotient = x / 10;
unsigned long long adjacent = quotient % 10;
if (remainder == adjacent) {
return false;
}
x = quotient;
remainder = adjacent;
}
return true;
}
int main(int argc, char *argv[]) {
char *pointer;
unsigned long long n = strtoull(argv[1], &pointer, 10) + 1;
while (true) {
if (valid(n)) {
printf("%llu\n", n);
break;
}
n++;
}
return 0;
}
> dc
10 10 ^ p
10000000000
> ./a.out 10000000000
10101010101
> dc
10 11 ^ p
100000000000
> ./a.out 100000000000
101010101010
>
#include <stdio.h>
#include <stdlib.h>
unsigned logTen(unsigned long long number) {
unsigned power = 0;
while (number >= 10) {
power += 1;
number /= 10;
}
return power;
}
unsigned long long expTen(unsigned n) {
unsigned long long product = 1;
while (n > 0) {
product *= 10;
n -= 1;
}
return(product);
}
unsigned long long next_no_adjacent(unsigned long long number) {
number += 1;
unsigned power = logTen(number);
while (power > 0) {
unsigned long long multiplier = expTen(power);
unsigned long long digit = (number / multiplier) % 10;
unsigned long long adjacent_multiplier = expTen(power - 1);
unsigned long long adjacent_digit = (number / adjacent_multiplier) % 10;
while (digit == adjacent_digit) {
number = ((number + adjacent_multiplier) / adjacent_multiplier) * adjacent_multiplier;
digit = (number / multiplier) % 10;
adjacent_digit = (number / adjacent_multiplier) % 10;
}
--power;
}
return number;
}
int main(int argc, char *argv[]) {
char *pointer;
unsigned long long n = strtoull(argv[1], &pointer, 10);
printf("%llu\n", next_no_adjacent(n));
return 0;
}
> dc
10 19 ^ p
10000000000000000000
> time ./a.out 10000000000000000000
10101010101010101010
0.001u 0.001s 0:00.00 0.0% 0+0k 0+0io 0pf+0w
>
#include <bits/stdc++.h>
#define ll unsigned long long
using namespace std;
bool noDec(string str,int i,ll num){
str[i+1]--;
return stoll(str)>num;
}
ll num;
string str;
int main() {
cin>>num;
str=to_string(num+1);
for(int i=0; i<str.length()-1; i++){
if(str[i]==str[i+1]){
if(str[i]!='9'){
if(noDec(str,i,num)&&str[i]!='0')str[i+1]--;
else str[i+1]++;
}
else{
if(i==0){
str='1'+string(str.length(),'0');
}else{
str[i-1]++;
for(int j=i; j<str.length(); j++){
str[j]='0';
}
}
}i=0;
}
}
if(str[0]==str[1]&&str[0]=='9'){
str=string((ll)log10(num)+2,'1');
for(int i=1; i<str.length(); i+=2)
str[i]='0';
}
cout<<str;
}