2 seconds
256 megabytes
standard input
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 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.
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.
Print on the single line "YES" if n is
a nearly lucky number. Otherwise, print "NO" (without the quotes).
40047
NO
7747774
YES
1000000000000000000
NO
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".
很幸运看到了这道水题，立马把它A了，题意就是一个数所包含4或者7的数量如果是Lucky Number，那么就输出YES,反之，NO,水吧，重要在样例，一般看hint或note,这是最关键的；
<pre name="code" class="plain">#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=50;
char a[N];
int main()
{
int x,i,j;
while(~scanf("%s",a))
{
i=j=0;
x=strlen(a);
while(x--)
{
if(a[i]=='4'||a[i]=='7')
j++;
i++;
}
int f=0;
if(j==0)
f=1;
else
{
while(j)//其实这可以不这样写的，数据范围并不大，j如果符合肯定是个位数； {只需判断j是否等于4或7；
if(j%10!=4&&j%10!=7)
f=1;
j/=10;
}
} if(f)
printf("NO\n");
else
printf("YES\n");
}
return 0;
}