TZOJ 5094 Stringsobits(DP)

描述

Consider an ordered set S of strings of N (1 <= N <= 31) bits. Bits, of course, are either 0 or 1.

This set of strings is interesting because it is ordered and contains all possible strings of length N that have L (1 <= L <= N) or fewer bits that are `1'.

Your task is to read a number I (1 <= I <= sizeof(S)) from the input and print the Ith element of the ordered set for N bits with no more than L bits that are `1'.

输入

A single line with three space separated integers: N, L, and I.

输出

A single line containing the integer that represents the Ith element from the order set, as described.

样例输入

5 3 19

样例输出

10011

题意

N位的二进制串，从小到大排序，问'1'的个数<=L的第I个串。

题解

采用逐位确定法，如果要第N位填上'1'，那么前N-1位<=L的串的个数<=I。

那么如何确定前N-1位<=L的串的个数呢？

dp[i][j]代表第i位填了j个'1'的串的个数，

当第i位填'1'，那么就有dp[i-1][j-1]。

当第i位填‘0’，那么就有dp[i-1][j]。

代码

 #include<bits/stdc++.h>
 using namespace std;
 #define ll long long
 ll dp[][],I;
 int main()
 {
     int N,L;
     cin>>N>>L>>I;
     dp[][]=;
     for(int i=;i<=N;i++)
         for(int j=;j<=L;j++)
             if(j==)dp[i][j]=dp[i-][j];
             else if(j>i)dp[i][j]=dp[i][i];
             else if(j==i)dp[i][j]=dp[i][j-]+;
             else dp[i][j]=dp[i-][j]+dp[i-][j-];
     for(int i=N-;i>=;i--)
         if(dp[i][L]<I){cout<<"";I-=dp[i][L--];}
         else cout<<"";
     if(I!=)cout<<"";
     else cout<<"";
     return ;
 }