An Easy Problem

GitHub 跳转原题关系图返回列表

题意与原解析均从本地 OpenJudge 缓存迁移。

OJ: noi_openjudge

题目 ID: ch0406-1455

难度:未知

标签:

日期: 2026-07-30 23:01

题意

完整题面见同目录的 problem.md。

思路

代码

cpp

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45#include <cstdio>
#include <cstring>

int n;


//进行操作
void getAns(int x){

    int cnt = 0; //计算用

    //找到第一个 符合条件的1
    int i;
    for(i=0;i<32;i++){
        if(  (x &(1<<i)) !=0 && ( x & (1<<(i+1))) == 0){
            //fined
            x = ( x | (1<<(i+1))); //第i+1位置1

            int t = ~((1<<(i+1))-1); // 1111000这种样式
            x = x & t; //后面全置0

            t = (1<<cnt)-1;
            x = x | t;
            break;
        }
        else if((x &(1<<i)) !=0)
            cnt++;
    }

    printf("%d\n",x);

}

int main(){
    
    while(1){
        scanf("%d",&n);
        if( n == 0)  break;

        getAns(n);


    }
    return 0;
}

复杂度

总结

ch0406-1455 An Easy Problem

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 '1’s in whose binary form is the same as that in the binary form of I.For example, if “78” is given, we can write out its binary form, “1001110”. This binary form has 4 '1’s. The minimum integer, which is greater than “1001110” and also contains 4 '1’s, is “1010011”, i.e. “83”, so you should output “83”.

输入

One integer per line, which is I (1 <= I <= 1000000).A line containing a number “0” terminates input, and this line need not be processed.

输出

One integer per line, which is J.

样例输入

text

样例输出

text

来源

POJ Monthly,zby03

限制

时间限制：1000ms
内存限制：128MB