UVa 674 Coin Change "Memory Search"

Test instructions: Give 1,5,10,25,50 five kinds of coins, and then give N, ask how many different schemes can gather n

Write your own writing when the program number is always less (one-dimensional)

Later found that the search, to use two-dimensional to write

http://blog.csdn.net/keshuai19940722/article/details/11025971

This article says there will be a repetition of the face value, not quite understand

There is a preprocessing problem, after looking at the puzzle to write their own, it is customary to handle the DP array to write to the while loop inside, has been tle

And then I saw this.

Http://www.cnblogs.com/scau20110726/archive/2012/12/25/2832968.html

Because the topic does not say how many sets of data, if the processing of the DP array in the while loop, if you give a 10w group of data, it will definitely time out (equivalent to each time, to deal with the DP Array)

So just put the pretreatment on the outside.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <stack>6#include <vector>7#include <map>8#include <Set>9#include <queue>Ten#include <algorithm> One #definemod=1e9+7; A using namespacestd; -  -typedefLong LongLL; the Const intINF =0x7fffffff; - Const intmaxn=8005; -LL dp[maxn][ the]; - intcoin[5]={1,5,Ten, -, -}; +  -LL DFS (intSinti) { +     if(dp[s][i]!=-1)returnDp[s][i]; A      atdp[s][i]=0; -      for(intj=i;j<5&&s>=coin[j];j++) -Dp[s][i]+=dfs (S-coin[j],j); -      -     returnDp[s][i];  - } in  - intMain () { to     intN; +memset (dp,-1,sizeof(DP)); -          for(intI=0;i<5; i++) dp[0][i]=1; the      while(SCANF ("%d", &n)! =EOF) { *      $printf"%lld\n", DFS (N,0));Panax Notoginseng     } -     return 0; the}

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UVa 674 Coin Change "Memory Search"