第4届CCCC字符串大赛 L2-1特立独行的幸福

/*题解详见公众号文章,最底扫码关注*/
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
const int N = 1e6 + 7;
#define IO do{\
ios::sync_with_stdio(false);\
cin.tie(0);\
cout.tie(0);}while(0)
#define mem(a,b) memset(a,b,sizeof(a));
mapd;
int su[10005] = {1,1,0};
void isprime()
{
	for (int i = 2; i < 10005; i++)
		if (!su[i])
			for (int j = i*i; j < 10005; j += i)
				su[j] = 1;
}
int main()
{
	int n, m;
	cin >> n >> m;
	int b[10005], l[10005],c=0,dd[5];
	isprime();
	for (int i = n; i <= m; i++)
	{
		int a = i,cc=0,flag1=0;
		while (a != 1)
		{
			cc++;
			dd[0] = a / 10000;
			dd[1] = a / 1000 % 10;
			dd[2] = a / 100 % 10;
			dd[3] = a / 10 % 10;
			dd[4] = a % 10;
			a = dd[0] * dd[0] + dd[1] * dd[1] + dd[2] * dd[2] + dd[3] * dd[3] + dd[4] * dd[4];
			if (cc > 10)
			{
				flag1 = 1;
				break;
			}
		}
		if (!flag1)
		{
			l[c] = cc;
			b[c++] = i;
		}
	}
	for (int i = 0; i < c; i++)
	{
		int a = b[i], flag1 = 0;
		while (a != 1)
		{
			dd[0] = a / 10000;
			dd[1] = a / 1000 % 10;
			dd[2] = a / 100 % 10;
			dd[3] = a / 10 % 10;
			dd[4] = a % 10;
			a = dd[0] * dd[0] + dd[1] * dd[1] + dd[2] * dd[2] + dd[3] * dd[3] + dd[4] * dd[4];
			d[a]++;
		}
	}
	int flag1 = 0;
	for (int i = 0; i < c; i++)
	{
		int a = b[i];
		if (d.count(a) && (a >= n&& a <= m))
			continue;
		else
		{
			flag1++;
			cout << a << " ";
			if (su[a])
				cout << l[i] << endl;
			else
				cout << 2 * l[i] << endl;
		}
	}
	if (!flag1)
		cout << "SAD" << endl;
	return 0;
}

第4届CCCC字符串大赛 L2-1特立独行的幸福_第1张图片

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