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; }