1. 程式人生 > 實用技巧 >[CF從零單排#16]110A - Nearly Lucky Number

[CF從零單排#16]110A - Nearly Lucky Number

題目來源:http://codeforces.com/problemset/problem/110/A

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 47, 744, 4 are lucky and 5, 17, 467 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".

題目大意:

這題題目意思不是太清晰,真實的題意應該是這樣:4,7都是幸運數字,如果一個數所有數字都是幸運數字,那麼這個數是幸運數。一個數中,如果幸運數字的個數是幸運數,那麼這個數叫做類幸運數。問某個數是否為類幸運數。

題目分析:

題意清晰後,本題就是很簡單的模擬題。因為題目提示數字可能是longlong,那麼數字個數最多是64,因此存在的幸運數只有4,7,47。所以只需要統計有多少個幸運數,再判定它們是否等於4,7,47即可。

參考程式碼:

#include <bits/stdc++.h>
using namespace std;
int main(){
	long long x;
	cin >> x;
	int cnt = 0;
	while(x){
		int t = x%10;
		if(t==4 || t==7){
			cnt ++;
		}
		x /= 10;
	}
	if(cnt==4 || cnt==7 || cnt == 47)
		cout << "YES";
	else
		cout << "NO";
	return 0;
}