HDU - 5491 The Next 2015 ACM/ICPC Asia Regional Hefei Online

从D+1开始，对于一个数x，区间[x,x+lowbit(x))内的数字的二进制位上1的数量整体来说是单调不减的，因此可快速得出1在这个区间的取值范围。

每次判断一下有没有和[s1，s2]有没有交集，一旦发现解就贪心最小的一个。

复杂度是O(T*log(ans-D))

#include<bits/stdc++.h>
using namespace std;

inline int read()
{
char c; while(c=getchar(),c<'0'||c>'9');
int re = c-'0';
while(c=getchar(),c>='0'&&c<='9') re = re*10+c-'0';
return re;
}

#define lowbit(x) (x&-x)
int bc(long long x)
{
int re = 0;
while(x){
re += x&1;
x>>=1;
}
return re;
}

//bitset<64> bs;

#define cer(x) bs = x; cout<<bs<<endl;

//#define LOCAL
int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif
int T = read();
for(int ks = 1; ks <= T; ks++){
int D = read(), s1 = read(), s2 = read();
long long cur = D+1LL,lb = lowbit(cur);
int ct = bc(cur);
while(ct < s1 || ct > s2){
int ex = bc(lb-1);
if(s1 > ct+ex || s2 < ct) {
cur += lb;
lb = lowbit(cur);
ct = bc(cur);
}else {
int ad = max(ct,s1)-ct;
cur += (1<<ad)-1;
break;
}
}
printf("Case #%d: %I64d\n",ks,cur);
}
return 0;
}