【链接】 我是链接,点我呀:)
【题意】
【题解】
最多有64位 我们可以从l开始一直增大到r 怎么增大? 找到l的二进制表示当中0所在的位置 假设i这一位的0经过加法变成了1 那么我们再从低位到高位依次枚举那一位为1就好 然后把这个二进制转换成十进制 看看它是不是比ri来的小 如果是,则尝试更新最小值 a[i]=(int)一个long类型的数字%2 a[i]是Int 会出错..另解
从低位到高位依次让每个0bit都变成1bit
看看新变成的数字是否小于等于ri就好.
如果小于等于,那么就继续尝试让更高位的0变成1
因为是从低位到高位,因此每次增加1个0的代价肯定都是最小的
cin >>l>>r;
for(long long i=1;(l|i)<=r;i<<=1)l|=i;
cout <<l<<endl;
【代码】
import java.io.*;
import java.util.*;
public class Main {
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static int N = 50000;
static class Task{
int n,len;
long li,ri;
long Bits[];
public void solve(InputReader in,PrintWriter out) {
Bits = new long[100];
n = in.nextInt();
for (int ii = 1;ii <= n;ii++) {
li = in.nextLong();ri = in.nextLong();
long ans = li;int ans1 = 0;
for (int j = 63;j >= 1;j--){
Bits[j] = li%2;
li = li/2;
ans1 += Bits[j];
}
//out.println("ans="+ans);
//out.println("ans1="+ans);
long temp = ans;
int temp1 = ans1;
long now = 1;
for (int j = 63;j >= 1;j--) {
temp = temp - now*Bits[j];
temp1-=Bits[j];
if(Bits[j]==0) {
temp1 = temp1+1;
temp = temp + now;
//out.println("temp="+temp+" now = "+now);
if (temp<=ri) {
if (temp1>ans1) {
ans1 = temp1;
ans = temp;
}else if (temp1==ans1 && temp<ans) {
ans = temp;
}
}
long temp2 = 1,cur = 0;
for (int k = 1;k <= (63-j);k++) {
cur = cur + temp2;
if (temp + cur<=ri) {
if (temp1 + k > ans1) {
ans1 = temp1 + k;
ans = temp + cur;
}else if (temp1+k==ans1 && temp+cur<ans) {
ans = temp + cur;
}
}else break;
temp2 = temp2*2;
}
temp = temp - now;
temp1 = temp1-1;
}
now = now * 2;
}
out.println(ans);
//out.println(ans1);
}
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
tokenizer = null;
}
public String next(){
while (tokenizer==null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}