Codeforces刷题之路——110A Nearly Lucky Number

A. Nearly Lucky Number
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 477444 are lucky and 517467 are not.

Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number n is a nearly lucky number.

Input

The only line contains an integer n (1 ≤ n ≤ 1018).

Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.

Output

Print on the single line "YES" if n is a nearly lucky number. Otherwise, print "NO" (without the quotes).

Examples
input
40047
output
NO
input
7747774
output
YES
input
1000000000000000000
output
NO
Note

In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".

In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".

In the third sample there are no lucky digits, so the answer is "NO".



题目大意:本题判断的是nearly lucky number,而非lucky number,该数字中包含lucky number 4,7的个数,该个数是不是lucky number(即只有4 or 7 组成的数字)
解题思路:先计算一下该输入数字中含有lucky number的个数,在检测该个数是不是仅含有4 or 7 即可。


以下为解题代码(java实现)
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scanner = new Scanner(System.in);
		String string = scanner.nextLine();
		
		int count = 0;
		for(int i = 0;i < string.length();i++){
			if(string.charAt(i) == '4' || string.charAt(i) == '7'){
				count++;
			}
		}
		
		boolean flag = false;
		String string2 = String.valueOf(count);
		for(int i = 0;i < string2.length();i++){
			if(string2.charAt(i) == '4' || string2.charAt(i) == '7'){
				flag = true;
				continue;
			}else{
				flag = false;
				break;
			}
		}
		if(flag){
			System.out.println("YES");
		}else{
			System.out.println("NO");
		}
		scanner.close();
	}
}

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