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[32][32],I;
int main()
{
    int N,L;
    cin>>N>>L>>I;
    dp[0][0]=1;
    for(int i=1;i<=N;i++)
        for(int j=0;j<=L;j++)
            if(j==0)dp[i][j]=dp[i-1][j];
            else if(j>i)dp[i][j]=dp[i][i];
            else if(j==i)dp[i][j]=dp[i][j-1]+1;
            else dp[i][j]=dp[i-1][j]+dp[i-1][j-1];
    for(int i=N-1;i>=1;i--)
        if(dp[i][L]<I){cout<<"1";I-=dp[i][L--];}
        else cout<<"0";
    if(I!=1)cout<<"1";
    else cout<<"0";
    return 0;
}