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法律研究長度為n且兩半和相等的二進位制數的所有可能性？
我們將在這裡檢視n位二進位制數（n由使用者指定）的所有可能性，其中每半邊的和相同。例如，如果數字是10001，則10和01相同，因為它們的和相同，並且它們位於不同的半邊中。我們將在此處生成所有此類數字。
演算法
genAllBinEqualSumHalf(n, left, right, diff)
left和right最初為空，diff保持left和right之間的差
Begin if n is 0, then if diff is 0, then print left + right end if return end if if n is 1, then if diff is 0, then print left + 0 + right print left + 1 + right end if return end if if 2* |diff| <= n, then if left is not blank, then genAllBinEqualSumHalf(n-2, left + 0, right + 0, diff) genAllBinEqualSumHalf(n-2, left + 0, right + 1, diff-1) end if genAllBinEqualSumHalf(n-2, left + 1, right + 0, diff + 1) genAllBinEqualSumHalf(n-2, left + 1, right + 1, diff) end if End
示例
#include <bits/stdc++.h>
using namespace std;
//left and right strings will be filled up, di will hold the difference between left and right
void genAllBinEqualSumHalf(int n, string left="", string right="", int di=0) {
if (n == 0) { //when the n is 0
if (di == 0) //if diff is 0, then concatenate left and right
cout << left + right << " ";
return;
}
if (n == 1) {//if 1 bit number is their
if (di == 0) { //when difference is 0, generate two numbers one with 0 after left, another with 1 after left, then add right
cout << left + "0" + right << " ";
cout << left + "1" + right << " ";
}
return;
}
if ((2 * abs(di) <= n)) {
if (left != ""){ //numbers will not start with 0
genAllBinEqualSumHalf(n-2, left+"0", right+"0", di);
//add 0 after left and right
genAllBinEqualSumHalf(n-2, left+"0", right+"1", di-1);
//add 0 after left, and 1 after right, so difference is 1 less
}
genAllBinEqualSumHalf(n-2, left+"1", right+"0", di+1); //add 1 after left, and 0 after right, so difference is 1 greater
genAllBinEqualSumHalf(n-2, left+"1", right+"1", di); //add 1 after left and right
}
}
main() {
int n = 5;
genAllBinEqualSumHalf(n);
}輸出
100001 100010 101011 110011 100100 101101 101110 110101 110110 111111
更新於： 31-Jul-2019
190次瀏覽
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