Codeforces 431 D. Random Task 数位dp单调性

xiaoxiao2021-02-28  115

很巧妙的单调性......

n,n+1,n+2,.....2*n-2

   n+1,n+2,.....2*n-2,2*n-1,2*n

中间一段是相同的,n和2*n里的1是一样多的所以只有2*n-1不一样.....这是满足单调性的(monotone)

然后就是数位DP了.....

D. Random Task time limit per test  1 second memory limit per test  256 megabytes input  standard input output  standard output

One day, after a difficult lecture a diligent student Sasha saw a graffitied desk in the classroom. She came closer and read: "Find such positive integer n, that among numbers n + 1n + 2, ..., 2·n there are exactly m numbers which binary representation contains exactlyk digits one".

The girl got interested in the task and she asked you to help her solve it. Sasha knows that you are afraid of large numbers, so she guaranteed that there is an answer that doesn't exceed 1018.

Input

The first line contains two space-separated integers, m and k (0 ≤ m ≤ 10181 ≤ k ≤ 64).

Output

Print the required number n (1 ≤ n ≤ 1018). If there are multiple answers, print any of them.

#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> using namespace std; long long dp[105][105]; long long m; int k; int num[105]; long long DFS(int pos, int sum1,bool limit) { if(pos == -1) { if(sum1==k) return 1; else return 0; } if(limit == 0 && dp[pos][sum1] != -1) return dp[pos][sum1]; long long sum = 0; int up = limit ? num[pos] : 1; for( int i=0; i<=up; i++ ) { int tp=sum1; if(i==1) tp++; sum += DFS(pos - 1, tp , limit && i == num[pos]); } if(limit == 0) dp[pos][sum1] = sum; return sum; } long long Solve(long long n) { int kk = 0; while(n) { num[kk++] = n % 2; n /= 2; } return DFS(kk-1,0,1); } long long bin() { long long l=1,r=1e18; while(l<r) { long long mid=(l+r)/2; long long tans=Solve(2*mid)-Solve(mid); if(tans==m) return mid; else if(tans>m) r=mid-1; else l=mid+1; } } int main() { while(~scanf("%lld%d",&m,&k)) { memset(dp,-1,sizeof(dp)); long long ans=bin(); printf("%lld\n",ans); } }

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