Let L denote the number of 1s in integer D's binary representation. Given two integers S1 and S2, we call D a XHY number if S1≤L≤S2.
With a given D, we would like to find the next XHY number Y, which is JUST larger than D. In other words, Y is the smallest XHY number among the numbers larger than D. Please write a program to solve this problem.
输入要求
The first line of input contains a number T indicating the number of test cases (T≤100000).
Each test case consists of three integers D, S1, and S2, as described above. It is guaranteed that 0≤D<2^28 and D is a XHY number.
输出要求
For each test case, output a single line consisting of “Case #X: Y”. X is the test case number starting from 1. Y is the next XHY number.
测试数据示例
输入
3
11 2 4
22 3 3
15 2 5
输出
Case #1: 12
Case #2: 25
Case #3: 17
大水题，开始以为会TLE，结果暴力搞一下位运算就过了，主要是题目有难懂。
题目意思大概如下：
L代表整数D的二进制位为1的个数。譬如D=3,L=2。
假如s1<=L<=s2，则D就是XHY数。现在就是要你找出大于D的最小XHY数。
题目明白了之后就是很裸的按位与(&)操作。
上代码了.
C code
#include <stdio.h>
int slove(int d,int s1,int s2){
int next;
long end = << ;
for(int i=d+;i<=end;i++){
int bitCount = ;
for(int j=;j<;j++){
if( (i & (<<j)) > ) bitCount++;
}
if(bitCount>=s1 && bitCount <= s2){
next=i;
break;
}
}
return next;
}
int main(){
int n;
int d;
int s1;
int s2;
int c = ;
while(scanf("%d",&n)==){
for(int i=;i<n;i++){
scanf("%d",&d);
scanf("%d",&s1);
scanf("%d",&s2);
int ret = slove(d,s1,s2);
printf("Case #%d: %d\n",c++,ret);
}
}
}