HDU 1058 humble numbers

Now that we classify this question into the dynamic planning questions, we start by thinking about it with the idea of dynamic planning.

There was no breakthrough in the morning.

Define four pseudo pointers to store the subscript, indicating the subscript of the smallest number that can be multiplied by 2, 3, 5, and 7 in the known ugly number.

The above sentence should be well understood.

That is to say, DP [P1] * 2 can get a new ugly number, and the other three are like this.

But what we need is the next ugly number in the largest known ugly number, that is, the minimum value of the new ugly number.

That is, min (DP [P1] * 2, DP [P2] * 3, DP [P3] * 5, DP [P4] * 7 );

 

In addition, it should be noted that in dp (), when the ugly numbers generated by one or two are the same as the minimum values, the corresponding "Pointer" must also be auto-incremented by 1.

For example, when I = 5, P1 = 3, P2 = 2, DP [P1] * 2 = DP [P2] * 3 = 6;

If you only execute ++ P1 and do not execute ++ P2, the ugly number 9 is missing.

 

Finally, I spoke about the egg pain output ..

When can I ignore the question, I think about DP question a by myself ..

 

 1 //#define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 using namespace std; 7  8 int dp[5850]; 9 10 inline int min(int a, int b, int c, int d)11 {12     a = min(a, b);13     c = min(c, d);14     return min(a, c);15 }16 17 void DP(void)18 {19     int i;20     int p1 = 1, p2 = 1, p3 = 1, p4 = 1;21     dp[1] = 1;22     for(i = 2; i <= 5842; ++i)23     {24         dp[i] = min(dp[p1] * 2, dp[p2] * 3, dp[p3] * 5, dp[p4] * 7);25         if(dp[i] == dp[p1] * 2)26             ++p1;27         if(dp[i] == dp[p2] * 3)28             ++p2;29         if(dp[i] == dp[p3] * 5)30             ++p3;31         if(dp[i] == dp[p4] * 7)32             ++p4;33     }34 } 35 36 int main(void)37 {38     #ifdef LOCAL39         freopen("1058in.txt", "r", stdin);40     #endif41 42     DP();43     int n;44     while(scanf("%d", &n) && n)45     {46         if(n % 10 == 1 && n % 100 != 11)47             printf("The %dst humble number is ", n);48         else if (n % 10 == 2 && n % 100 != 12)49         {50             printf("The %dnd humble number is ", n);51         }52         else if (n % 10 == 3 && n % 100 != 13)53         {54             printf("The %drd humble number is ", n);55         }56         else57             printf("The %dth humble number is ", n);58 59         printf("%d.\n", dp[n]);60     }61     return 0;62 }

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