给定三个整数P ， Q和R ，任务是生成一个具有P，Q和R对的连续字符对的二进制字符串，分别具有和0、1和2。
例子：
Input: P = 1, Q = 2, R = 2
Output: 111001
Explanation:
Substrings “00”, “10”, “01”, and “11” have sums 0, 1, 1, and 2 respectively.
Thus, the following set of substrings { {“11”}, {“11”}, {“10”}, {“00”}, {“01”} } satisfy the given constraints. Hence, the string formed by the substrings is 111001.
Input: P = 3, Q = 1, R = 0
Output: 10000
方法：为了解决此问题，我们需要遵循以下步骤：
- 如果P和R不为零，则至少存在一对和为1的连续字符。因此，如果提供的Q为0，则在这种情况下无法形成这样的字符串。因此，返回-1 。
- 如果Q是零，只有P的一个和R是非零的，则附加0 P + 1次，如果P是非零或附加1个R +如果R是非零1周时间。
- 如果它们都不为零：
- 分别追加0和1 P +1和Q +1次。
- Q – 1交替追加0和1 。
下面是上述方法的实现：
C++
// C++ program to generate a binary
// string with given frequencies
// of sums of consecutive
// pair of characters
#include
using namespace std;
// A Function that generates
// and returns the binary string
string build_binary_str(int p,
int q, int r)
{
// P: Frequency of consecutive
// characters with sum 0
// Q: Frequency of consecutive
// characters with sum 1
// R: Frequency of consecutive
// characters with sum 2
string ans = "";
// If no consecutive
// character adds up to 1
if (q == 0) {
// Not possible if both P
// and Q are non - zero
if (p != 0 && r != 0) {
return "-1";
}
else {
// If P is not equal to 0
if (p) {
// Append 0 P + 1 times
ans = string(p + 1, '0');
}
else {
// Append 1 R + 1 times
ans = string(r + 1, '1');
}
}
}
else {
// Append "01" to satisfy Q
for (int i = 1; i <= q + 1; i++) {
if (i % 2 == 0) {
ans += '0';
}
else {
ans += '1';
}
}
// Append "0" P times
ans.insert(1, string(p, '0'));
// Append "1" R times
ans.insert(0, string(r, '1'));
}
return ans;
}
// Driver Code
int main()
{
int p = 1, q = 2, r = 2;
cout << build_binary_str(p, q, r);
return 0;
} Java
// Java program to generate a binary
// String with given frequencies
// of sums of consecutive
// pair of characters
class GFG{
// A Function that generates
// and returns the binary String
static String build_binary_str(int p,
int q, int r)
{
// P: Frequency of consecutive
// characters with sum 0
// Q: Frequency of consecutive
// characters with sum 1
// R: Frequency of consecutive
// characters with sum 2
String ans = "";
// If no consecutive
// character adds up to 1
if (q == 0)
{
// Not possible if both P
// and Q are non - zero
if (p != 0 && r != 0)
{
return "-1";
}
else
{
// If P is not equal to 0
if (p > 0)
{
// Append 0 P + 1 times
ans = Strings(p + 1, '0');
}
else
{
// Append 1 R + 1 times
ans = Strings(r + 1, '1');
}
}
}
else
{
// Append "01" to satisfy Q
for (int i = 1; i <= q + 1; i++)
{
if (i % 2 == 0)
{
ans += '0';
}
else
{
ans += '1';
}
}
// Append "0" P times
ans = ans.substring(0, 1) + Strings(p, '0') + ans.substring(1);
// Append "1" R times
ans = Strings(r, '1') + ans;
}
return ans;
}
static String Strings(int p, char c)
{
String ans = "";
for (int i = 0; i < p; i++)
ans += c;
return ans;
}
// Driver Code
public static void main(String[] args)
{
int p = 1, q = 2, r = 2;
System.out.print(build_binary_str(p, q, r));
}
}
// This code is contributed by shikhasingrajputPython3
# Python3 program to generate a binary
# string with given frequencies
# of sums of consecutive
# pair of characters
# A Function that generates
# and returns the binary string
def build_binary_str(p, q, r):
# P: Frequency of consecutive
# characters with sum 0
# Q: Frequency of consecutive
# characters with sum 1
# R: Frequency of consecutive
# characters with sum 2
ans = ""
# If no consecutive
# character adds up to 1
if (q == 0):
# Not possible if both P
# and Q are non - zero
if (p != 0 and r != 0):
return "-1"
else:
# If P is not equal to 0
if (p):
# Append 0 P + 1 times
temp = ""
for i in range(p + 1):
temp += '0'
ans = temp
else:
# Append 1 R + 1 times
temp = ""
for i in range(r + 1):
temp += '1'
ans = temp
else:
# Append "01" to satisfy Q
for i in range(1, q + 2):
if (i % 2 == 0):
ans += '0'
else:
ans += '1'
# Append "0" P times
temp = ""
for i in range(p):
temp += '0'
st = ""
st += ans[0]
st += temp
for i in range(1, len(ans)):
st += ans[i]
ans = st
# Append "1" R times
temp = ""
for i in range(r):
temp += '1'
ans = temp + ans
return ans
# Driver Code
if __name__ == '__main__':
p = 1
q = 2
r = 2
print(build_binary_str(p, q, r))
# This code is contributed by Surendra_GangwarC#
// C# program to generate a binary
// String with given frequencies
// of sums of consecutive
// pair of characters
using System;
class GFG{
// A Function that generates
// and returns the binary String
static String build_binary_str(int p,
int q, int r)
{
// P: Frequency of consecutive
// characters with sum 0
// Q: Frequency of consecutive
// characters with sum 1
// R: Frequency of consecutive
// characters with sum 2
String ans = "";
// If no consecutive
// character adds up to 1
if (q == 0)
{
// Not possible if both P
// and Q are non - zero
if (p != 0 && r != 0)
{
return "-1";
}
else
{
// If P is not equal to 0
if (p > 0)
{
// Append 0 P + 1 times
ans = Strings(p + 1, '0');
}
else
{
// Append 1 R + 1 times
ans = Strings(r + 1, '1');
}
}
}
else
{
// Append "01" to satisfy Q
for (int i = 1; i <= q + 1; i++)
{
if (i % 2 == 0)
{
ans += '0';
}
else
{
ans += '1';
}
}
// Append "0" P times
ans = ans.Substring(0, 1) +
Strings(p, '0') +
ans.Substring(1);
// Append "1" R times
ans = Strings(r, '1') + ans;
}
return ans;
}
static String Strings(int p, char c)
{
String ans = "";
for (int i = 0; i < p; i++)
ans += c;
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int p = 1, q = 2, r = 2;
Console.Write(build_binary_str(p,
q, r));
}
}
// This code is contributed by 29AjayKumar输出：
111001
时间复杂度： O(P + Q + R)