没有重复的最少位数
least number of digits without duplicates
我是 C++ 菜鸟。我有一个我放入 Vector 中的数字列表。所有数字都是 9 位整数并且是唯一的。我想知道可用于唯一标识集合中每个数字的最少位数(从右边开始)是多少。现在只有 6 个数字,但这个列表可能会增长到数千个。到目前为止,我已经发布了我的代码(无法正常工作。)
EDIT 输出如下...
digit is 1
digit is 1
digit is 1
RUN FINISHED; exit value 0; real time: 0ms; user: 0ms; system: 0ms
这主要是一个学习练习。请慷慨而明确地提出您的意见和解决方案。
#include <iostream>
#include <vector>
#include <fstream>
#include <string>
#include <cstdlib>
#include <algorithm>
using namespace std;
int main() {
//declare stream variable and load vector with values
ifstream myfile("mydata.txt");
vector<int> myVector;
int num;
while (myfile >> num) {
myVector.push_back(num);
}
//sort and squack if there is a duplicate.
std::sort(myVector.begin(), myVector.end());
for (int i = 0; i < (myVector.size() - 1); i++) {
if (myVector.at(i) == myVector.at(i + 1)) {
printf("There are duplicate student numbers in the file");
exit(EXIT_FAILURE);
}
}
//if it get here, then there are no duplicates of student numbers
vector<int> newv;
int k = 1;
bool numberFound = false;
bool myflag = false;
while (numberFound == false) {
//loop through original numbers list and add a digit to newv.
for (int j = 0; j < myVector.size(); ++j) {
newv.push_back(myVector.at(j) % (10^k));
}
sort(newv.begin(), newv.end());
for (int i = 0; i < (newv.size() - 1); i++) {
if (newv.at(i) == newv.at(i + 1)) {
//there is a duplicate for this digit. Set flag.
myflag = true;
}
if (myflag == false) {
numberFound = true;
cout << "digit is " << k << endl;
} else {
k++;
}
}
}
// for (int i = 0; i < myVector.size(); i++) {
// cout << "||" << myVector.at(i) << "||" << endl;
// }
//
// for (int i = 0; i < newv.size(); i++) {
// cout << "---" << newv.at(i) << "---" << endl;
// }
return 0;
}
检查下面的代码。
#include <iostream>
#include <vector>
#include <fstream>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <math.h>
using namespace std;
int main() {
//declare stream variable and load vector with values
ifstream myfile("mydata.txt");
vector<int> myVector;
int num;
while (myfile >> num) {
myVector.push_back(num);
}
//sort and squack if there is a duplicate.
std::sort(myVector.begin(), myVector.end());
for (int i = 0; i < (myVector.size() - 1); i++) {
if (myVector.at(i) == myVector.at(i + 1)) {
printf("There are duplicate student numbers in the file");
exit(EXIT_FAILURE);
}
}
//if it get here, then there are no duplicates of student numbers
vector<int> newv;
int k = 1;
bool numberFound = false;
bool myflag = false;
int p = 1;
while (numberFound == false) {
//loop through original numbers list and add a digit to newv.
newv.clear();
p = p * 10;
for (int j = 0; j < myVector.size(); ++j) {
newv.push_back(myVector[j] % p);
}
sort(newv.begin(), newv.end());
myflag = false;
for (int i = 0; i < (newv.size() - 1); i++) {
if ( newv[i] == newv[i+1]) {
//there is a duplicate for this digit. Set flag.
myflag = true;
break;
}
}
if (myflag == true){
k ++;
}else{
numberFound = true;
cout << "digit is " << k << endl;
break;
}
}
return 0;
}
示例输入:
123451789
123456687
125456789
123456780
输出:
digit is 4
我是 C++ 菜鸟。我有一个我放入 Vector 中的数字列表。所有数字都是 9 位整数并且是唯一的。我想知道可用于唯一标识集合中每个数字的最少位数(从右边开始)是多少。现在只有 6 个数字,但这个列表可能会增长到数千个。到目前为止,我已经发布了我的代码(无法正常工作。)
EDIT 输出如下...
digit is 1
digit is 1
digit is 1
RUN FINISHED; exit value 0; real time: 0ms; user: 0ms; system: 0ms
这主要是一个学习练习。请慷慨而明确地提出您的意见和解决方案。
#include <iostream>
#include <vector>
#include <fstream>
#include <string>
#include <cstdlib>
#include <algorithm>
using namespace std;
int main() {
//declare stream variable and load vector with values
ifstream myfile("mydata.txt");
vector<int> myVector;
int num;
while (myfile >> num) {
myVector.push_back(num);
}
//sort and squack if there is a duplicate.
std::sort(myVector.begin(), myVector.end());
for (int i = 0; i < (myVector.size() - 1); i++) {
if (myVector.at(i) == myVector.at(i + 1)) {
printf("There are duplicate student numbers in the file");
exit(EXIT_FAILURE);
}
}
//if it get here, then there are no duplicates of student numbers
vector<int> newv;
int k = 1;
bool numberFound = false;
bool myflag = false;
while (numberFound == false) {
//loop through original numbers list and add a digit to newv.
for (int j = 0; j < myVector.size(); ++j) {
newv.push_back(myVector.at(j) % (10^k));
}
sort(newv.begin(), newv.end());
for (int i = 0; i < (newv.size() - 1); i++) {
if (newv.at(i) == newv.at(i + 1)) {
//there is a duplicate for this digit. Set flag.
myflag = true;
}
if (myflag == false) {
numberFound = true;
cout << "digit is " << k << endl;
} else {
k++;
}
}
}
// for (int i = 0; i < myVector.size(); i++) {
// cout << "||" << myVector.at(i) << "||" << endl;
// }
//
// for (int i = 0; i < newv.size(); i++) {
// cout << "---" << newv.at(i) << "---" << endl;
// }
return 0;
}
检查下面的代码。
#include <iostream>
#include <vector>
#include <fstream>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <math.h>
using namespace std;
int main() {
//declare stream variable and load vector with values
ifstream myfile("mydata.txt");
vector<int> myVector;
int num;
while (myfile >> num) {
myVector.push_back(num);
}
//sort and squack if there is a duplicate.
std::sort(myVector.begin(), myVector.end());
for (int i = 0; i < (myVector.size() - 1); i++) {
if (myVector.at(i) == myVector.at(i + 1)) {
printf("There are duplicate student numbers in the file");
exit(EXIT_FAILURE);
}
}
//if it get here, then there are no duplicates of student numbers
vector<int> newv;
int k = 1;
bool numberFound = false;
bool myflag = false;
int p = 1;
while (numberFound == false) {
//loop through original numbers list and add a digit to newv.
newv.clear();
p = p * 10;
for (int j = 0; j < myVector.size(); ++j) {
newv.push_back(myVector[j] % p);
}
sort(newv.begin(), newv.end());
myflag = false;
for (int i = 0; i < (newv.size() - 1); i++) {
if ( newv[i] == newv[i+1]) {
//there is a duplicate for this digit. Set flag.
myflag = true;
break;
}
}
if (myflag == true){
k ++;
}else{
numberFound = true;
cout << "digit is " << k << endl;
break;
}
}
return 0;
}
示例输入:
123451789
123456687
125456789
123456780
输出:
digit is 4