UVa 674 Coin Change (classic DP)

Coin Change

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld &%llu

Suppose there is 5 types of coins:50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money.

For example, if we had one cents, then we can make changes with one 10-cent coin and one 1-cent coin, one 5-cent coins and One 1-cent coin, one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there is four ways of making changes for one cents with the above coins. Note that we count this there is one of the making change for zero cent.

Write a program to find the total number of different ways of making changes for all amount of money in cents. Your program should is able to handle up to 7489 cents.

Input

The input file contains any number of lines, and each one consisting of a number for the amount of money in cents.

Output

For each input line, output a line containing the number of different ways of making changes with the above 5 types of COI Ns.

Sample Input

1126

Sample Output

413

Test instructions: There are 1,5,10,25,50 five kinds of coins, give a number, ask a few ways of pooling to make up this number.

Analysis: Classic DP problem ... can be recursive or memory search ... This problem test instructions and HDU2069 exactly the same, but the same code is not submitted, often time out, or see the idea of the great God to turn.

1#include <cstdio>2#include <cstring>3 Const intMAXN =8000;4 Const intcoin[5] = {1,5,Ten, -, -};5 intN;6 Long Longdp[maxn][5];7 8 Long LongSolveintIints)9 {Ten     if(Dp[s][i]!=-1) One         returnDp[s][i]; Adp[s][i]=0; -      for(intj=i;j<5&& s>=coin[j];j++) -Dp[s][i]+=solve (j,s-coin[j]); the     returnDp[s][i]; - } -  - intMain () + { -memset (dp,-1,sizeof(DP)); +      for(intI=0;i<5; i++) Adp[0][i]=1; at      while(~SCANF ("%d",&N)) -printf"%lld\n", Solve (0, N)); -     return 0; -}

UVa 674 Coin Change (classic DP)