Theit is no problemm is easy...

来源：蜘蛛抓取(WebSpider) 时间：2011-10-25 05:23 标签： it is no problem
he problem is very easy. They can work it out. (合并) The problem is easy __ for __ to work out.
The problem is easy eough for them to work out.
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扫描下载二维码i think the problem is very easy.think后是什么成分
I think接的是宾语从句,即一个句子作THINK 的宾语.
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i think后面加的是个句子
what do you think i should do? 经常是do you think/suppose/consider/the problem i think is very easy. the problem is very easy.看,没有
是宾语从句 省略了“that ”
扫描下载二维码4发现相似题POJ 2453 An Easy Problem [解题报告] Java
POJ 2453 An Easy Problem [解题报告] Java
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An Easy Problem
问题描述 :
As we known, data stored in the computers is in binary form. The problem we discuss now is about the positive integers and its binary form.
Given a positive integer I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of *s in whose binary form is the same as that in the binary form of I.
For example, if + is given, we can write out its binary form, &#″. This binary form has 4 *s. The minimum integer, which is greater than &#″ and also contains 4 *s, is &#″, i.e. +, so you should output +.
One integer per line, which is I (1 <= I <= 1000000).
A line containing a number &#43; terminates input, and this line need not be processed.
One integer per line, which is J.
import java.util.S
public class Main {
* ?峰?涓?釜Integer瀵瑰????杩??涓???釜??
* @param i
* @return sum
public static int getSum(int i) {
int sum = 0;
while (i > 0) {
i = i &#038; (i - 1);
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
int i = sc.nextInt();
if (i == 0)
int sum1 = getSum(i);
for (int j = i + 1;; ++j) {
int sum2 = getSum(j);
if (sum2 == sum1) {
System.out.println(j);
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