Envious Exponents

Posted 2021-01-01 rhythm-

问题 E: Envious Exponents

时间限制: 1 Sec  内存限制: 128 MB
提交: 321  解决: 53
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题目描述

Alice and Bob have an integer N. Alice and Bob are not happy with their integer. Last night they went to a cocktail party and found that another couple had the exact same integer! Because of that they are getting a new integer.

Bob wants to impress the other couple and therefore he thinks their new integer should be strictly larger than N.
Alice herself is actually fond of some speci?c integer k. Therefore, Alice thinks that whatever integer they pick, it should be possible to write it as a sum of k distinct powers of 2.

Bob is also a cheapskate, therefore he wants to spend as little money as possible. Since the cost of an integer is proportional to its size, he wants to get an integer that is as small as possible.

输入

• A single line containing two integers N and k, with 1 ≤ N ≤ 10¹⁸ and 1 ≤ k ≤ 60.

输出

Output M, the smallest integer larger than N that can be written as the sum of exactly k distinct powers of 2.

样例输入

1 2

样例输出

3

题意：找出大于n且二进制中恰好有k个1的最小整数。

代码：

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <deque>
#include <map>
#include <unordered_map>
#include <set>
#include <time.h>
#define P(n,f) cout<<n<<(f?‘
‘:‘ ‘)
#define range(i,a,b) for(auto i=a;i<=b;++i)
#define LL long long
#define ULL unsigned long long
#define elif else if
#define itrange(i,a,b) for(auto i=a;i!=b;++i)
#define rerange(i,a,b) for(auto i=a;i>=b;--i)
#define fill(arr,tmp) memset(arr,tmp,sizeof(arr))
#define IOS ios::sync_with_stdio(false);cin.tie(0)
using namespace std;
LL n,k;
LL bit[65],tail,cnt;
void init(){
    fill(bit,0);
    cin>>n>>k;
}
void solve(){
    while(cin>>n>>k) {
        fill(bit,0);cnt=0;tail=0;
        while (n) {
            bit[++tail] = (n & 1);
            cnt += (n & 1);
            n >>= 1;
        }
        if (cnt >= k) {
            int tmp, pos = 0;
            for (tmp = 0; tmp < cnt and bit[tail - tmp]; ++tmp);
            while (cnt > k) {
                if (!bit[++pos]) continue;
                bit[pos] = 0;
                --cnt;
            }
            if (tmp >= cnt) {
                fill(bit, 0);
                bit[++tail] = 1;
                cnt = 1;
            } else {
                range(i, pos+1, tail-1)
                    if(bit[i]) {
                        if (not bit[i + 1]) {
                            bit[i + 1] = 1;
                            bit[i] = 0;
                            break;
                        }
                        else{
                            bit[i]=0;
                            --cnt;
                        }
                    }
            }
        }
        int pos = 0;
        while (cnt < k)
            if (not bit[++pos]) {
                bit[pos] = 1;
                ++cnt;
            }
        LL ans = 0, add = 1;
        range(i, 1, 64) {
            ans += bit[i] * add;
            add <<= 1;
        }
        P(ans, 1);
    }
}
int main() {
    //init();
    solve();
    return 0;
}

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