Codeforces 110A-Nearly Lucky Number(实现)

最新推荐文章于 2020-12-10 11:11:28 发布

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最新推荐文章于 2020-12-10 11:11:28 发布

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分类专栏： Codeforces 文章标签： codeforces

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本文介绍了一种判断Nearly Lucky Number的方法，即一个数中幸运数字4和7的数量是否为幸运数。通过解析输入数字并计数幸运数字出现的次数，再判断该次数是否为幸运数来完成判断。

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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 test(s)

input
40047

output
NO

input
7747774

output
YES

input
1000000000000000000

output
NO题意：给出幸运数字的概念是该数字仅含4和7. 然后给出Nearly Lucky Number 的定义：对于一个数n，它每位上4或7的个数如果是幸运数，输出YES，否则输出NO。#include <iostream>#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <set>
#include <map>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
const int INF=0x3f3f3f3f;
#define LL long long
int main()
{
    LL n;char s[22],c[22];
    while(cin>>n)
	{
		sprintf(s,"%lld",n);
		int cnt=0;
		for(int i=0;i<strlen(s);i++)
		{
			if(s[i]=='4'||s[i]=='7')
				cnt++;
		}
		sprintf(c,"%d",cnt);
		int flag=1;
		for(int i=0;i<strlen(c);i++)
		{
			if(c[i]!='4'&&c[i]!='7')
			{
				flag=0;
				break;
			}
		}
		if(flag)
			puts("YES");
		else
			puts("NO");
	}
	return 0;
}