【题目】Find the missing number in the( ), make the statement ture. (Hint: T here are more than one answer(1)()+7+10=20(2)()+6-5=10(3)()+4=10-((4)18-()=()+10 相关知识点: 试题来源: 解析 【解析】(1)(3)+7+10=20.(2)(9)+6-5=10.(3)(1)+4=10-(5).(答案不...
Find the missing number in the sequence:3,6,11,20,37,___,135.Answer:___. 相关知识点: 试题来源: 解析 70 70 因为6=3×2-0;11=6×2-1;20=11×2-2;37=20×2-3;所以下个数应该为:37×2-4=70;验证:70×2-5=135正好是最后一个数;故答案为:70 反馈 收藏 ...
Find the missing number in the sequence:3,6,11,20,37,___,135.Answer:___. 答案 因为6=3×2-0;11=6×2-1;20=11×2-2;37=20×2-3;所以下个数应该为:37×2-4=70;验证:70×2-5=135正好是最后一个数;故答案为:70 结果三 题目 Find the missing number in the sequence:3,6,11,20,...
// C program to find the missing number in the array#include <stdio.h>intmain() {intarr[]={1,2,3,5,6};intsize=0;inti=0;intmissing=0; size=sizeof(arr)/sizeof(arr[0]); missing=(size+1)*(size+2)/2;for(i=0; i<size; i++) missing=missing-arr[i]; printf("Missing numbe...
Find the missing number in the ( ), make the statement ture. (Hint: There are more than one answer) (1)( )+7+10=20 (2)( )+6−5=10 (3)( )+4=10−( ) (4)18−( )=( )+10 相关知识点: 试题来源: 解析 (1)(3)+7+10=20. (2)(9)+6−5=10. (3)(1)+4=10...
Given an array containsNnumbers of 0 ..N, find which number doesn't exist in the array. Example GivenN=3and the array[0, 1, 3], return2. Challenge Do it in-place with O(1) extra memory and O(n) time. 这道题是LeetCode上的原题,请参见我之前的博客Missing Number 丢失的数字。那...
Find the missing number in the pattern. 根据规律,找出横线上的数字. 1,3,5,7, ,12,16,18,23,25….相关知识点: 试题来源: 解析 10 1+2=3,3+2=5,5+2=7,12+4=16,16+2=18,18+5=23,23+2=25⋯.. 由此可得规律,第奇数个数与第偶数个数之间的差值规律为:2,2,3,2,4,2,5,2,6,2...
【题目】Find the missing number in the pattern.根据规律,找出横线上的数字100,94,88,82,70,64 相关知识点: 试题来源: 解析 【解析】76 结果一 题目 【题目】Find the missing number in the pattern.根据规律,找出横线上的数字1,2,4,714,16,23,25, 答案 【解析】9...
Find the Missing Number Given an array containsNnumbers of 0 ..N, find which number doesn't exist in the array. Example GivenN=3and the array[0, 1, 3], return2. Challenge Do it in-place with O(1) extra memory and O(n) time. ...
// Find the missing number in a given array intgetMissingNumber(intarr[],intn) { // the actual size is `n+1` since a number is missing from the array intm=n+1; // get a sum of integers between 1 and `n+1` inttotal=m*(m+1)/2; ...