Pat class B 1016 C language

1016. part A+B (15)

Positive integer A Of “D_A（ by 1 An integer ） part ” Defined by A All in D_A A new integer composed of P_A. for example ： Given A = 3862767,D_A = 6, be A Of “6 part ”P_A yes 66, because A There is 2 individual 6.

Now given A、D_A、B、D_B, Please write a program to calculate P_A + P_B.

Input format ：

The input is given in turn on a line A、D_A、B、D_B, Space between , among 0 < A, B < 10¹⁰.

Output format ：

Output in one line P_A + P_B Value .

sample input 1：

3862767 6 13530293 3

sample output 1：

399

sample input 2：

3862767 1 13530293 8

sample output 2：

0

Ideas ： Use two for Loop to count the number of characters to be counted in the two strings , So we can calculate two numbers , Add one

One 、 Starting variable

1.input1 and input2 These are two variables that accept strings

2.a and b The characters to be found

3.countOfA and countOfB Is the number of occurrences corresponding to two characters

4.A and B Is the last corresponding number

Two 、 operation

1. Accept two strings and two characters

2. use for Loop through two strings , Count the number of occurrences of characters to be searched

3. To calculate the A and B

4. Add and A and B Output

3、 ... and 、 Code

#include "stdio.h"
#include "string.h"
#include "math.h"
typedef char string[10];

int main()
{
	string input1,input2;
	char a,b;// Two characters to find  
	int A = 0, B = 0;// Finally used to calculate sum A and B
	scanf("%s %c %s %c",input1, &a, input2, &b);
	char * p;
	int countOfA = 0, countOfB = 0;
	// Statistics a Number of occurrences  
	int i = 0;
	for(i = 0; i < strlen(input1); i++)
	{
		if(input1[i] == a)
		countOfA++;
	}
	// Statistics b Number of occurrences  
	for(i = 0; i < strlen(input2); i++)
	{
		if(input2[i] == b)
		countOfB++;
	} 
	// Separate calculation A、B 
	for(i = 0; i < countOfA; i++)
	{
		A += (a - '0') * pow(10, i);
	}
	for(i = 0; i < countOfB; i++)
	{
		B += (b - '0') * pow(10, i);
	}
	printf("%d",A + B);
	return 0;
}