A. Nearly Lucky Number
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”.
问题链接：CodeForces - 110A
问题简述：判断一个小于十的十八次方的数即十八个数字中，幸运数字的个数是否为幸运数字，幸运数字为4或7.第一行输入这串数字，第二行判断。
问题分析：1、’这串数字为十的十八次方，正好可以用%lld接收这串数字然后每次除十来判断该数字是否为幸运数字。2、可以用数组接收十八个数字然后用循环判断该数字是否为幸运数字。3、也可用getchar（）一个数字一个数字地接收并处理。用count统计幸运数字数量并判断其是否为幸运数字4 7。
程序说明：用getchar（）一个一个接收数字，减少了内存占用，加快了运行速率（？）
AC通过的C语言程序如下：
#include<stdio.h>
int main()
{
char n,count=0;
n=getchar();
while(n!='\n')
{
if(n=='4'||n=='7')
{
count++;
}
n=getchar();
}
if(count==7||count==4)
{
printf("YES\n");
}else printf("NO\n");
return 0;
}