牛牛的好朋友羊羊在纸上写了n+1个整数，羊羊接着抹除掉了一个整数，给牛牛猜他抹除掉的数字是什么。牛牛知道羊羊写的整数神排序之后是一串连续的正整数，牛牛现在要猜出所有可能是抹除掉的整数。例如：
10 7 12 8 11 那么抹除掉的整数只可能是9
5 6 7 8 那么抹除掉的整数可能是4也可能是9
[编程题]连续整数
- 热度指数：4318 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 32M，其他语言64M
- 算法知识视频讲解
输入描述:
输入包括2行：
第一行为整数n(1 <= n <= 50)，即抹除一个数之后剩下的数字个数
第二行为n个整数num[i] (1 <= num[i] <= 1000000000)
输出描述:
在一行中输出所有可能是抹除掉的数,从小到大输出,用空格分割,行末无空格。如果没有可能的数，则输出mistake
示例1
输入
2 3 6
输出
mistake
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt(); //抹除一个数之后剩下的数字个数
int[] num = new int[n];
for (int i = 0; i < n; i ++) {
num[i] = sc.nextInt();
}
//如果只有一个数字，注意保证输入正整数
if (n == 1) {
if (num[0] > 1) {
System.out.println(num[0] - 1 + " " + (num[0] + 1));
return;
} else {
System.out.println((num[0] + 1));
return;
}
}
//多个数字，先排序
Arrays.sort(num);
//如果有重复数字，则mistake
for (int i = 0; i < n-1; i ++) {
if (num[i] == num[i+1]) {
System.out.println("mistake");
}
}
//现在没有了重复数字，那么可以根据最大数num[n-1]和最小数num[0]的差值来比较，
//这样去判断，中间数字是连续的，还是缺了一个，还是缺了多个
if (num[n-1] - num[0] > n) {
System.out.println("mistake");
} else if (num[n-1] - num[0] == n) {
for (int i = 0; i < n-1; i ++) {
if (num[i] + 2 == num[i+1]) {
System.out.println(num[i] + 1);
}
}
} else {
if (num[0] > 1) {
System.out.println(num[0] - 1 + " " + (num[n-1] + 1));
} else {
System.out.println((num[n-1] + 1));
}
}
}
}
发表于 2017-03-12 14:08:09
回复(3)
#include<iostream>
using namespace std;
int main() {
int n;
cin>>n;
int min = 1000000001;
int max = 0;
int res = 0;
int num = 0;
for(int i = 0; i < n; i++) {
cin>>num;
if(num < min) min = num;
if(num > max) max = num;
res ^= num;
}
for(int i = min; i <= max; i++) {
res ^= i;
}
if(res == 0) {
if(min - 1> 0)
cout<<min-1<<" "<<max+1;
else cout<<max+1;
}
else if(res <= min || res >= max) {
cout<<"mistake"<<endl;
}
else cout<<res;
}
编辑于 2017-03-08 17:31:26
回复(11)
思路：对串排序。然后用最大值减最小值，如果差为n，则涂抹的是中间数字；如果差为n-1，则涂抹两边数字；其他，mistake。先判断是否有重复数字。注意最小值为1的情况。
#include <set>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
int main()
{
int n;
vector<int> t_s;
cin >> n;
for (int i = 0; i < n; i++)
{
int tt;
cin >> tt;
t_s.push_back(tt);
}
sort(t_s.begin(), t_s.end());
set<int> s;
for (int j = 0; j < n; j++){
s.insert(t_s[j]);
}
if (s.size() == n)//判断是否有数重复
{
if (t_s[n - 1] - t_s[0] == n - 1)
{
if(t_s[0]-1!=0){
cout << t_s[0] - 1 << ' ' << t_s[n - 1]
+ 1 << endl;}
else cout <<t_s[n - 1] + 1 << endl;
}
else if (t_s[n - 1] - t_s[0] == n){
for (int k = t_s[0]; k <= t_s[n - 1]; k++)
{
if (s.count(k) == 0) {
cout << k << endl;
return 0;
}
}
}
else cout << "mistake" << endl;
}
else cout << "mistake" << endl;
return 0;
}
编辑于 2017-03-08 15:20:50
回复(0)
#我来用python写一个，见识一下python的强大
def get_the_lose_one(data): data_ordered = sorted(data,lambda x,y:x-y) outcome= [] for index,data in enumerate(data_ordered): if index == len(data_ordered)-2: break if data+1 != data_ordered[index+1]: outcome.append(data+1) if len(outcome) >= 1: print 'mistake' elif not len(outcome): print [data_ordered[0]-1,data_ordered[len(data_ordered)-1]+1] else: print outcome[0]
发表于 2017-09-21 21:53:27
回复(1)
/*
参考的小风筝的答案，不过那个原答案不能通过所有测试用例，我加了个判断
*/
import java.util.Arrays;
import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int[] num=new int[n];
for(int i=0;i<n;i++){
num[i]=sc.nextInt();
}
if(n==1){
if(num[0]>1){
System.out.println(num[0]-1+" "+(num[0]+1));
}else{
System.out.println((num[0]+1));
}
}
Arrays.sort(num);
for(int i=0;i<n-1;i++){
if (num[i] == num[i+1]) {
System.out.println("mistake");
}
}
if(num[n-1]-num[0]>n){
System.out.println("mistake");
}else if(num[n-1]-num[0]==n){
for (int i = 0; i < n-1; i ++) {
if (num[i] + 2 == num[i+1]) {
System.out.println(num[i] + 1);
}
}
}else{
if (num[0] > 1&&n!=1) {
System.out.println(num[0] - 1 + " " + (num[n-1]
+ 1));
} else if(num[0]!=1&&n!=1){
System.out.println((num[n-1] + 1));
}
}
if(num[0]==1&&num[n-1]==n&&n!=1){//如果n==1&&num[0]!=1
System.out.println((num[n-1] + 1));
}
}
}
发表于 2017-03-24 16:17:43
回复(0)
/*
5种情况
1、内缺：
i. 缺1个 直接输出结果
ii. 缺多个(间接) mistake
iii.缺多个(连续) mistake
2、 外缺：
i.第一个元素==1 数列右边缺
ii.第一个元素>1 数列左右都有可能缺
*/
#include <iostream>
#include <algorithm>
using namespace std;
const int MAX_ = 50;
int arr[MAX_];
int main() {
int n,count = 0,tmp;
cin >> n;
for(int i=0; i<n; i++) {
cin >> arr[i];
}
sort(arr,arr+n);
for(int i=1; i<n; i++) {
if(arr[i]-arr[i-1] == 2) {
tmp = arr[i] - 1;
count++;
continue;
} else if(arr[i]-arr[i-1] > 2) { // 缺多个(连续)
cout << "mistake" << endl;
return 0;
}
}
if(count == 1) { // 缺1个
cout << tmp << endl;
return 0;
} else if(count > 1) { // 缺多个(间接)
cout << "mistake" << endl;
return 0;
}
if(arr[0] > 1) { // 数列左右都有可能缺
cout << arr[0]-1 << " " << arr[n-1]+1 << endl;
} else { // 数列右边缺
cout << arr[n-1]+1 << endl;
}
return 0;
}
发表于 2017-03-23 22:38:52
回复(1)
while True: list_get_str = input('请输入一串整数（以空格分隔）').split(' ') list_get = list(map(int, list_get_str)) list_get.sort() print(list_get) list_length = len(list_get) count = 0 count_list = [] mistake = 0 for i in range(list_length-1): if list_get[i] + 1 == list_get[i+1]: continue elif list_get[i] + 2 == list_get[i + 1]: count += 1 count_list.append(list_get[i]) elif (list_get[i] >= list_get[i+1])&nbs***bsp;(list_get[i] + 2) < list_get[i+1]: print('mistake') mistake = 1 break if mistake == 0: if len(count_list) == 0 : print(list_get[0]-1) print(list_get[-1] + 1) elif len(count_list) == 1: print(count_list[0]) elif len(count_list) >= 2: print('mistake')
编辑于 2020-08-26 22:28:16
回复(0)
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner in = new Scanner(System.in);
int length = in.nextInt();
int[] numbers = new int[length];
for (int i = 0; i<length; i++){
numbers[i]=in.nextInt();
}
if(length==1) {
if(numbers[0]==1) {
System.out.println(2);
}
else {
System.out.print((numbers[0]-1) +" "+ (numbers[0]+1));
}
}
else {
Arrays.sort(numbers);
ArrayList<Integer> index = new ArrayList<Integer>();
for(int i = 0; i<length-1; i++){
if(numbers[i+1]-numbers[i]==0||numbers[i+1]-numbers[i]>2){
System.out.println("mistake");
return;
}
if(numbers[i+1]-numbers[i]==2) {
index.add(i);
}
}
if(index.size()==0) {
if(numbers[0]==1) {
System.out.print((numbers[length-1]+1));
}
else {
System.out.print((numbers[0]-1) +" "+ (numbers[length-1]+1));
}
}
else if (index.size()==1) {
System.out.println(numbers[index.get(0)]+1);
}
else {
System.out.println("mistake");
}
}
}
} 那些年躲不过的坑！！！！
发表于 2020-04-03 08:34:52
回复(0)
#include <iostream>
#include <vector>
using namespace std;
class Solution
{
public:
void Swap(int& a, int& b)
{
a ^= b ^= a ^= b;
}
// 选择排序
void SortVec(vector<int>& vec)
{
int len = vec.size();
if (len == 1)
{
return;
}
int minPos;
for (size_t i = 0; i < len; i++)
{
minPos = i;
for (size_t j = i + 1; j < len; j++)
{
if (vec[minPos] > vec[j])
{
minPos = j;
}
}
if (minPos != i)
{
Swap(vec[minPos], vec[i]);
}
}
}
void Func1(vector<int>& vec, int Num)
{
// 1、先排序
SortVec(vec);
int gap = vec[Num - 1] - vec[0];
// 涂抹中间数字
if ( gap == Num)
{
for (size_t i = 0; i < Num; i++)
{
int gapTmp = vec[i + 1] - vec[i];
if (gapTmp == 0) // 如果有重复数字
{
cout << "mistake";
return;
}
if (gapTmp != 1)
{
cout << vec[i] + 1;
return;
}
}
}
else if ( gap == (Num - 1) ) // 涂抹两端数字
{
if (vec[0] == 1) cout << vec[Num - 1] + 1; // 如果最小值为1, 涂抹的是最右边(1 <= num[i] <= 1000000000)
else
{
cout << vec[0] - 1 << " " << vec[Num - 1] + 1;
}
}
else
{
cout << "mistake";
}
}
};
int main()
{
Solution s;
vector<int> vec;
int Num;
cin>>Num;
for (size_t i = 0; i < Num; i++)
{
int num;
cin >> num;
vec.push_back(num);
}
s.Func1(vec, Num);
return 0;
}
发表于 2019-09-07 22:53:01
回复(1)
#include <stdlib.h>
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
using namespace std;
int partition(int a[], int low, int high) {
int pivotPos = low;
int pivot = a[low];
int tmp;
for (int i = low + 1; i <= high; ++i) {
if (a[i] <= pivot) {
++pivotPos;
if (i != pivotPos) {
tmp = a[i];
a[i] = a[pivotPos];
a[pivotPos] = tmp;
}
}
}
a[low] = a[pivotPos];
a[pivotPos] = pivot;
return pivotPos;
}
void Qsort(int a[], int low, int high) {
if (low < high) {
int pivotPos = partition(a, low, high);
Qsort(a, low, pivotPos - 1);
Qsort(a, pivotPos + 1, high);
}
}
int main() {
#include <stdlib.h>
#include <math.h>
#include <stdio.h>
using namespace std;
int partition(int a[], int low, int high) {
int pivotPos = low;
int pivot = a[low];
int tmp;
for (int i = low + 1; i <= high; ++i) {
if (a[i] <= pivot) {
++pivotPos;
if (i != pivotPos) {
tmp = a[i];
a[i] = a[pivotPos];
a[pivotPos] = tmp;
}
}
}
a[low] = a[pivotPos];
a[pivotPos] = pivot;
return pivotPos;
}
void Qsort(int a[], int low, int high) {
if (low < high) {
int pivotPos = partition(a, low, high);
Qsort(a, low, pivotPos - 1);
Qsort(a, pivotPos + 1, high);
}
}
int main() {
int n;
scanf("%d", &n);
int* a = new int[n];
for (int i = 0; i < n; ++i)
scanf("%d", a + i);
Qsort(a, 0, n - 1);
int count = 0;
int pos;
for (int i = 0; i < n-1; ++i) {
if (a[i + 1] - a[i] == 2) {
pos = i;
++count;
}
else if (a[i + 1] - a[i] > 2)
count += a[i + 1] - a[i] - 1;
if (count > 1)
break;
}
if (count > 1) {
printf("mistake");
}
else if (count == 0) {
if (a[0] - 1 != 0) {
printf("%d ", a[0] - 1);
}
printf("%d", a[n - 1] + 1);
}
else {
printf("%d", a[pos] + 1);
}
delete[]a;
return 0;
}
发表于 2019-05-03 12:00:44
回复(0)
import java.util.Arrays; import java.util.Scanner; public class Main { public static void main(String args[]) { Scanner sc = new Scanner(System.in); int n,k,count=0; n=sc.nextInt(); int[] a=new int[n]; for(int i=0;i<n ;i++) a[i]=sc.nextInt(); Arrays.sort(a); k=a[0];//把排序后数组的第一个数赋值给一个常量 for(int i=1;i<n;i++)//计算有多少个相邻数字差值大于1的情况存在，并把数量计入count { if(a[i]==a[i-1]+1) continue; else count++; } for(int i=0;i<n;i++,k++){//开始遍历数组，开始k的值和啊a[0]相同，每次k++寻找差值大于1的数 if(a[i]!=k&&a[i]==k+1&&count==1)//如果数组和k的值不相等，且整个数组中只有这一种情况存在(即count==1)，则这个值为所求值 { System.out.print(k); System.exit(0); } else if(a[i]!=k&&a[i]!=k+1||count>1)//如果数组和k的值不相同，但是k和数组的差值大于1，则这种情况没有可能的数，并且如果count>1，也不存在 { System.out.print("mistake"); System.exit(0); } }//整个数组是一个有序数列，如果所有数之间的差值都为1，则进行如下判断 if(a[0]==1) {//如果这个数组第一个数为1，则只可能是末尾丢失，所求数位数组最后一个数+1 System.out.print(a[n-1]+1); } else System.out.print(a[0]-1+" "+(a[n-1]+1));//如果数组第一个数不是1，根据题意判断有两个值，即头-1尾+1 } }
发表于 2019-02-28 14:54:12
回复(0)
num=input()
list=[int(i) for i in input().split()]list=sorted(list)
set=set(list)
for i in list:
if max(set)-min(set)>len(set):
print('mistake')
break
if i+1 not in list:
if i==list[-1] :
if list[0]==1 :
print(i+1)
break
print(str(list[0]-1)+" "+str(i+1))
break
print(i+1)
break
发表于 2018-08-08 10:57:39
回复(0)
// Swift 4.0
let n = Int(readLine()!)!
let line = readLine()!let list = (line.split(separator: " ").map { Int($0)! }).sorted { $0 < $1 }
var result = [Any]()
var allContinuous = true
for i in 0 ..< n-1 {
let a = list[i+1] - list[i]
if a == 1 {
continue
}
else if a == 2 {
if result.count == 0 {
result.append(list[i]+1)
allContinuous = false
}
else {
result = ["mistake"]
break
}
}
else {
result = ["mistake"]
break
}
}
if allContinuous && result.count == 0 {
if list[0]-1 > 0 {
result.append(list[0]-1)
}
result.append(list[n-1]+1)
}
for i in 0 ..< result.count {
print(result[i], terminator: i == result.count-1 ? "\n" : " ")
}
发表于 2018-03-08 17:17:07
回复(0)
public class Main {
public static void main(String[] args) {
java.util.Scanner sc = new java.util.Scanner(System.in);
int n = sc.nextInt();
int[] numbers = new int[n];
for (int i = 0; i < n; i++) {
numbers[i] = sc.nextInt();
}
guess(numbers);
}
private static void guess(int[] numbers) {
if (numbers.length == 0) {
mistake();
} else if (numbers.length == 1) {
print(numbers[0] - 1, numbers[0] + 1);
} else {
int min = numbers[0], max = numbers[0], sum = numbers[0], first = numbers[0];
boolean[] bitArr = new boolean[numbers.length * 2 + 1];
bitArr[numbers.length] = true;
for (int i = 1; i < numbers.length; i++) {
sum += numbers[i];
//以first为参照物，将数字对应位数置为true
int index = numbers[i] - first;
if ((index > 0 && index <= numbers.length) || (index < 0 && -index <= numbers.length)) {
bitArr[numbers.length - index] = true;
} else if (index != 0) {//数字间距离过大，无法连续
mistake();
return;
}
if (numbers[i] > max) {
max = numbers[i];
}
if (numbers[i] < min) {
min = numbers[i];
}
}
int actualCount = numbers.length;
//求不重复数字个数
int uniqueCount = 0;
for (int i = 0; i < bitArr.length; i++) {
if (bitArr[i]) {
uniqueCount++;
}
}
if (actualCount != uniqueCount) {
mistake();
return;
}
//期望的数字个数
int expectedCount = max - min + 1;
//缺少的数字个数
int lackCount = expectedCount - uniqueCount;
if (lackCount == 0) {//数列连续
print(min - 1, max + 1);
} else if (lackCount == 1) {//中间缺少一个
print(0, (max + min) * (numbers.length + 1) / 2 - sum);
} else {//缺少多个
mistake();
}
}
}
private static void mistake() {
System.out.println("mistake");
}
private static void print(int i, int j) {
if (i > 0) {
System.out.println(i + " " + j);
} else {
System.out.println(j);
}
}
}
编辑于 2018-01-26 11:21:01
回复(0)
a = raw_input()
n = int(a)
b = raw_input()
c = b.split(' ')
num = map(int, c)
num = sorted(num)
ll = []
for i in range(n-1):
if num[i]+1 != num[i+1]:
ll.append(i)
if len(ll)==0:
if num[0]==1:
result = str(num[n-1]+1)
else:
result = str(num[0]-1)+' '+str(num[n-1]+1)
elif len(ll)==1:
result = str(num[ll[0]]+1)
else:
result = 'mistake'
print result
n = int(a)
b = raw_input()
c = b.split(' ')
num = map(int, c)
num = sorted(num)
ll = []
for i in range(n-1):
if num[i]+1 != num[i+1]:
ll.append(i)
if len(ll)==0:
if num[0]==1:
result = str(num[n-1]+1)
else:
result = str(num[0]-1)+' '+str(num[n-1]+1)
elif len(ll)==1:
result = str(num[ll[0]]+1)
else:
result = 'mistake'
print result
发表于 2018-01-09 22:50:24
回复(0)
#include <iostream>
#include <cmath>
#include <string>
#include <algorithm>
using namespace std;
int main(){
int n,m,count=0,mis=0;
cin >> n;
int num[50]={0};
int res[50]={0};
for(int i=0; i<n; i++){
cin >> num[i];
//cout <<num[i];
}
sort(num,num+n);
for(int i=0; i<n-1; i++){
if(num[i]+1 != num[i+1]){
count++;
}
}
if(count>1){
cout << "mistake";
return 0;
}
for(int i=0; i<n; i++){
if(num[i]+2 == num[i+1]){
cout << num[i] + 1;
return 0;
}else if(num[i]+2<num[i+1]){
cout << "mistake";
}
}
if(num[0]==1){
cout << num[n-1]+1;
}else{
cout << num[0]-1 <<" "<< num[n-1]+1;
}
return 0;
}
发表于 2017-12-20 16:54:07
回复(0)
int main(int argc, const char * argv[]) {
int n,i,j,temp=0,count=0;
bool ismistake=false;
cin>>n;
int num[n],array[4]={0,0,0,0};
for(i=0;i<n;i++){
cin>>num[i];
}
for(i=0;i<n;i++){
for(j=i+1;j<n;j++){
if(num[j]<num[i]){
temp = num[i];
num[i] = num[j];
num[j] = temp;
}
}
}
for(i=0;i<n-1;i++){
if((num[i+1]-num[i])>2){
cout<<"mistake"<<endl;
ismistake = true;
break;
}
if(num[i+1]-num[i]==2){
array[count] = num[i]+1;
count ++;
}
}
if(ismistake == false){
if(count == 0){
if(num[0]>1) cout<<num[0]-1<<" "<<num[n-1]+1;
else cout<<num[n-1]+1;
}
else if(count>=2) cout<<"mistake"<<endl;
else if(count == 1) cout<<array[0]<<endl;
}
return 0;
}
发表于 2017-11-06 20:29:12
回复(0)
import string if __name__ == '__main__': input_size = string.atoi(raw_input()) input_str = raw_input() input_int_list = [string.atoi(a) for a in input_str.split()] input_int_list.sort() if input_int_list[-1] - input_int_list[0] + 1 > input_size + 1: print 'mistake' if input_int_list[-1] - input_int_list[0] + 1 == input_size: if 1 != input_int_list[-1]: print str(input_int_list[0] - 1) + ' ' + str(input_int_list[-1] + 1) exit() print str(input_int_list[-1] + 1) for idx in range(len(input_int_list)-1): if input_int_list[idx] + 1 != input_int_list[idx+1]: print str(input_int_list[idx] + 1)
您的代码已保存
答案错误:您提交的程序没有通过所有的测试用例
case通过率为70.00%
测试用例:
3
1 3 5
对应输出应该为:
mistake
你的输出为:
mistake
答案错误:您提交的程序没有通过所有的测试用例
case通过率为70.00%
测试用例:
3
1 3 5
对应输出应该为:
mistake
你的输出为:
mistake
编辑于 2017-11-02 15:52:36
回复(0)
def find_miss(nums):
_min = float("+inf")
_max = float("-inf")
actual_total = 0
for k in nums:
flag = _max > _min
if flag:
if k > _max:
_max = k
elif k < _min:
_min = k
else:
if k > _max:
_max = k
if k < _min:
_min = k
actual_total = actual_total + k
predict_total = (_min + _max) * (_max - _min + 1)/2
miss = predict_total - actual_total
if miss == 0:
print _min - 1, 'or', _max + 1
elif miss > _max:
print "mistake"
else:
print miss
if __name__ == '__main__':
find_miss([5,6,7,8])
find_miss([5,6,9,10])
find_miss([5,6,8,9])
编辑于 2017-10-24 16:32:55
回复(0)
问题信息
通过挑战的用户
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2022-09-09 14:21:21
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2022-07-27 16:37:06
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2022-06-23 00:09:33 -
2022-06-10 14:54:38
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2022-05-09 17:42:53
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