CodeForces 110A Nearly Lucky Number
2017-03-05 09:21
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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).
Sample Input
Input
Output
Input
Output
Input
Output
Hint
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".
只需要判断字符串中‘4’和‘7’的总数是4个或者7个就行。
代码如下:
#include<stdio.h>
#include<string.h>
int main()
{
char a[20];
int i,n=0;
scanf("%s",a);
int len=strlen(a);
for(i=0;i<len;i++)
{
if(a[i]=='4'||a[i]=='7') n++;
}
if(n==4||n==7) printf("YES\n");
else printf("NO\n");
return 0;
}
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).
Sample Input
Input
40047
Output
NO
Input
7747774
Output
YES
Input
1000000000000000000
Output
NO
Hint
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".
只需要判断字符串中‘4’和‘7’的总数是4个或者7个就行。
代码如下:
#include<stdio.h>
#include<string.h>
int main()
{
char a[20];
int i,n=0;
scanf("%s",a);
int len=strlen(a);
for(i=0;i<len;i++)
{
if(a[i]=='4'||a[i]=='7') n++;
}
if(n==4||n==7) printf("YES\n");
else printf("NO\n");
return 0;
}
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