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UVA - 674 - Coin Change (背包型DP)

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UVA - 674

Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu

 Status

Description

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Suppose there are 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 have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, two 5-cent coins and one 1-cent coin, one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent.


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

Input 

The input file contains any number of lines, 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 coins.

Sample Input 

11
26

Sample Output 

4
13



Miguel Revilla
2000-08-14

Source

Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 3. Problem Solving Paradigms :: Dynamic Programming :: Coin Change - Classical
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 5. Dynamic Programming 
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms :: Dynamic Programming :: Coin Change (CC)
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Dynamic Programming :: Coin Change (CC)

 Status







思路:就是简单的动态规划问题,自底向上求解,从小往大循环


注意这里可以不用long long,最大值刚好没有超过int最大值


AC代码(未打表,2000+ms):

#include <cstdio> 
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int dp[10005]; 
int coin[5] = {1, 5, 10, 25, 50};

int main() {
	int n;
	while(scanf("%d", &n) != EOF) {
		memset(dp, 0, sizeof(dp));
		dp[0] = 1; 
		for(int i = 0; i < 5; i++) {
			for(int j = coin[i]; j <= n; j++) {
				dp[j] += dp[j - coin[i]];
			}
		}
		
		printf("%d\n", dp[n]);
	}
	return 0;
}






AC代码(打表,35ms,这里数据范围不大而又有很多组,可以打个表,速度更快):

#include <cstdio> 
#include <cstring>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;

LL dp[10005]; 	//dp[i]代表有i个cent的时候可以有的方案数 
int coin[5] = {1, 5, 10, 25, 50};

void init() {
	dp[0] = 1;
	for(int i = 0; i < 5; i++) {
		for(int j = coin[i]; j < 7500; j++) {
			dp[j] += dp[j - coin[i]];
		}
	}
}

int main() {
	init(); 
	int n;
	while(scanf("%d", &n) != EOF) {
		printf("%lld\n", dp[n]);
	}
	return 0;
}









UVA - 674 - Coin Change (背包型DP)

原文:http://blog.csdn.net/u014355480/article/details/44623239

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