USTC OJ-1011 Problem of Deck (combinatorial mathematics, recursive relationship)

1. Title Description

According to test instructions, we can know:

Put a ruler on the edge of the table (length L), in order to keep it from falling, extend the maximum length of the desktop edge of 1/2l.

On the basis of the above, and then add a ruler, in order to keep it not falling, extend the edge of the ruler below the maximum length of 1/4l.

In addition to a ruler, in order to keep it from falling, extend the edge of the ruler below the maximum length of 1/6l.

Go down in turn ...

2. Algorithm design

By the above description, we can get a recursive relationship :

When the n ruler is accumulated, the length of the edge of the table extends

Length (0) = 0

Length (n) = Length (n-1) + 1/(2n), n≥1.

3. AC Code

1#include <stdio.h>2 #defineN 1000003 DoubleTable[n];//Table[i] I cards the length of the Extend4 5 voidPrintintn);6 7 intMain ()8 {9     intI, N; Ten     //Take table Onetable[0] =0; A      for(i =1; i < N; i++ ) -     { -Table[i] = table[i-1] +1.0/ (2*i); the     } -  -printf"# Cards overhang\n"); -      while(EOF! = scanf ("%d", &N)) +     { - print (n); +     } A } at  - voidPrintintN) - { -printf"%5d%.3lf\n", N, Table[n]); -}

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USTC OJ-1011 Problem of Deck (combinatorial mathematics, recursive relationship)