- Time Limit:
- 1000ms
- Memory Limit:
- 65536kB
- Description
- 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". - Input
- 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. - Output
- One integer per line, which is J.
- Sample Input
-
1
2
3
4
78
0
- Sample Output
-
2
4
5
8
83
- Source
- POJ Monthly,zby03
- 题意：给定多组测试数据，每个数据一个正整数N，如果把N换算成二进制，如果有x个1，如78的二进制有4个1，那么求一个数，使得该数比x大，且换算成二进制的话也有x个1，求这个数的最小值。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstdlib>
4 #include<algorithm>
5 #include<cmath>
6 #include<cstring>
7 using namespace std;
8 int N;
9 int a[100];
10 int binum[100];
11 inline void change(){
12 memset(a,0,sizeof(a));
13 memset(binum,0,sizeof(binum));
14 int tmp=N;
15 int now=1;
16 while(tmp/2>0){
17 a[now]=tmp%2;
18 tmp/=2;
19 now++;
20 }
21 a[now]=1;
22 a[0]=now;
23 }
24 inline void move(){
25 int i=1;
26 int cnt=0;
27 while(a[i]!=1||a[i+1]!=0){
28 if(a[i]==1) cnt++;
29 i++;
30 }
31 a[i+1]=1;
32 for(int k=1;k<=i;k++) a[k]=0;
33 for(int k=1;k<=cnt;k++) a[k]=1;
34 }
35 inline void calc(){
36 int ANS=0;
37 int base=1;
38 for(int j=1;j<=a[0]+1;j++){
39 ANS+=a[j]*base;
40 base=base*2;
41 }
42 cout<<ANS<<endl;
43 }
44 int main(){
45 for(;;){
46 scanf("%d",&N);
47 if(N==0) break;
48 change();
49 move();
50 calc();
51 }
52 return 0;
53 }
