Y - 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：https://vjudge.net/contest/160731#problem/Y。
问题简述：判断一个长整型中数字4和数字7的和中是否只有数字7和数字4。对的话输出“yes”,不对的话输出“no”。
问题分析：利用取余、除法将输入的长整型中的每一位数储存到数组中，利用循环结构对数组中每一位数进行判断，计算共有多少个幸运数字。由于最多只能有18位数，因此和也只能是4和7。
程序说明：pow(float a,int b)函数;声明长整型long long n;利用取余和除法能将输入的长整型中的每一位数字分离出来。
AC通过的C语言程序如下：
#include
#include
#include <string.h>
#include
using namespace std;
int main()
{
int a[19];
int i;
long long n;
cin >> n;
if (n >= 1 && n <= pow(10.0,18))
{
for (i = 0; i < 19; i++)//用一个数字数组
{
a[i] = n % 10;
n /= 10;
}
int sum = 0;
for (i = 0; i < 19; i++)
{
if (a[i] == 4 || a[i] == 7)
{
sum++;
}
else
continue;
}
if (sum == 4 || sum == 7)
{
cout << "YES" << endl;
}
else
{
cout << "NO" << endl;
}
}
else
{
return 0;
}
}