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<''||c>'');
    int re = c-'';
    while(c=getchar(),c>=''&&c<='') re = re*+c-'';
    return re;
}
 
#define lowbit(x) (x&-x)
int bc(long long x)
{
    int re = ;
    while(x){
        re += x&;
        x>>=;
    }
    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 = ; 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-);
            if(s1 > ct+ex || s2 < ct) {
                cur += lb;
                lb = lowbit(cur);
                ct = bc(cur);
            }else {
                int ad = max(ct,s1)-ct;
                cur += (<<ad)-;
                break;
            }
        }
        printf("Case #%d: %I64d\n",ks,cur);
    }
    return ;
}