HDU 2069 Coin Change (full backpack)

Source: Internet
Author: User

Coin Change

Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 16592 Accepted Submission (s): 5656


Problem Descriptionsuppose 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, or both 5-cent coins And one 1-cent coin, or 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 coins.

Inputthe input file contains any number of lines, each one consisting of a number (≤250) for the amount of money in cent S.

Outputfor each input line, output a line containing the number of different ways of making changes with the above 5 types of coins.

Sample Input1126

Sample Output413

Authorlily

SOURCE Zhejiang University of Technology Network tryouts

Recommendlinle | We have carefully selected several similar problems for you:1171 1398 1085 1028 2152 The first time to write a complete backpack, the problem is very simple, but note that the number of coins can not be greater than 100 test instructions : Give you a 1,5,10,25,50 currency, and then randomly enter a number of money, asking how many (up to 100 coins) are attached to the method with these denominations just to make up the money:
1#include <iostream>2#include <cstdio>3#include <cstring>4 using namespacestd;5 intdp[101][7500];6 intMain ()7 {8     intcoin[5]= {1,5,Ten, -, -};9     inti,j,n,m,k;TenMemset (DP,0,sizeof(DP)); Onedp[0][0]=1; A      for(i=0; i<5; i++)//coins to use -     { -          for(k=1; k<= -; k++)//Number of coins the         { -              for(J=coin[i]; j<=7500; J + +) -             { -dp[k][j]+=dp[k-1][j-Coin[i]]; +             } -         } +  A     } at      while(~SCANF ("%d",&N)) -     { -         intans=0; -          for(i=0; i<= -; i++) -ans+=Dp[i][n]; -printf"%d\n", ans); in     } -     return 0; to}

HDU 2069 Coin Change (full backpack)

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