练习三 1004

概述：有一个数列，它的成员都是2，3，5，7的倍数，输入一个n，求出这个数列的第n个数。

思路：先列一个表，把这些数据全求出来，利用循环，设a,b,c,d=1，分别对他们*2,*3,*5,*7,取最小值入表，并将其+1，依次类推，最后输出所求值即可。

感想：以前做过的题，拿来想想就有思路了。

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
LL dp[5843];
LL min(LL a, LL b, LL c, LL d)
{
LL temp1 = a < b ? a : b;
temp1 = temp1 < c ? temp1 : c;
temp1 = temp1 < d ? temp1 : d;
return temp1;
}
int main()
{

dp[1] = 1;
int a = 1, b = 1, c = 1, d = 1;
for (int i = 2; i <= 5842; ++i)
{
dp[i] = min(dp[a] * 2, dp[b] * 3, dp[c] * 5, dp[d] * 7);
if (dp[i] == dp[a] * 2) a++;
if (dp[i] == dp[b] * 3) b++;
if (dp[i] == dp[c] * 5) c++;
if (dp[i] == dp[d] * 7) d++;
}
int n;
while (cin>>n&&n)
{
if (n % 100 != 11 && n % 10 == 1)printf("The %dst humble number is %lld.\n", n, dp
);
else
if (n % 100 != 12 && n % 10 == 2)printf("The %dnd humble number is %lld.\n", n, dp
);
else
if (n % 100 != 13 && n % 10 == 3)printf("The %drd humble number is %lld.\n", n, dp
);
else
printf("The %dth humble number is %lld.\n", n, dp
);
}
return 0;
}