CF431D Random Task

https://www.luogu.com.cn/problem/CF431D

首先$n$关于个数是单调的，$n$越大，个数越多
然后就可以二分+数位DP解决了

怎么证明是单调的呢?
考虑$[n+1,2n]和[n+2,2n+2]$
中间$[n+2,2n]$是一样的，然后可以发现$n+1$和$2n+2$ 1的个数也是一样的
所以把$n+1$后和$n$做差只剩下一个$2n+1$，所以必然是单调的

code:

#include<bits/stdc++.h>
#define ll long long
#define N 85
using namespace std;
ll f[N][N][2], a[N];
ll m, k;
ll dfs(int len, int cnt, int lim) {
    if(!len) return cnt == k;
    ll &ret = f[len][cnt][lim];
    if(ret != -1) return ret;
    ret = 0;
    int sx = lim? a[len] : 1;
    for(int i = 0; i <= sx; i ++) ret += dfs(len - 1, cnt + i, lim && (i == sx));
    return ret;
}
ll calc(ll x) {
    memset(f, -1, sizeof f);
    int len = 0;
    while(x) a[++ len] = x & 1, x >>= 1;
    return dfs(len, 0, 1);
}
int main() {
    scanf("%lld%lld", &m, &k);
    ll l = 0, r = 1e18;
    while(l + 1 < r) {
        ll mid = (l + r) >> 1;
        if(calc(mid * 2) - calc(mid) >= m) r = mid;
        else l = mid;
    }
    printf("%lld", r);
    return 0;
}