B - Nearly Lucky Number

标签：nat   nta   single   ecif   字符串处理。   else   为什么   repr   numbers   

Problem description

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

40047

NO

7747774

YES

1000000000000000000

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".

解题思路：简单字符串处理。如果输入的字符串中出现‘4‘或‘7‘的次数总和num刚好为4次或7次，则输出"YES"。为什么只有4或7呢？因为规定的字符串长度最长为19，而幸运数字（只由4或7组成）有4，7，47...，第三个数47比最长长度值19还大，即num最大为19，所以只需看num是否为4或7，是则输出"YES"，否则输出"NO"。

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     char str[20];int num=0;
 5     cin>>str;
 6     for(int i=0;str[i]!=‘\0‘;++i)
 7         if(str[i]==‘4‘ || str[i]==‘7‘)num++;
 8     if(num==4 || num==7)cout<<"YES"<<endl;
 9     else cout<<"NO"<<endl;
10     return 0;
11 }

B - Nearly Lucky Number

标签：nat   nta   single   ecif   字符串处理。   else   为什么   repr   numbers   

原文地址：https://www.cnblogs.com/acgoto/p/9136394.html