UVA 674 - Coin Change
2015-08-19 19:11
260 查看
UVA 674 - Coin Change
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
多重背包问题。不过我是用深搜+记忆化写的,
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
多重背包问题。不过我是用深搜+记忆化写的,
[code]#include<iostream> #include <algorithm> #include <cstdio> #include <string.h> #define N 8000 using namespace std; const int coin[5]={1, 5, 10, 25, 50}; int dp [5], n; int dfs(int remain, int i) { if (dp[remain][i] != -1) { return dp[remain][i]; } dp[remain][i] = 0; for (int j = i; j < 5 && coin[j] <= remain; j++) { dp[remain][i] += dfs( remain - coin[j], j); } return dp[remain][i]; } int main() { #ifndef ONLINE_JUDGE freopen("1.txt", "r", stdin); #endif int i, j; memset(dp, -1, sizeof(dp)); for(i = 0; i < 5; i++) { dp[0][i] = 1; } dfs(7500, 0); while(~scanf("%d", &n)) { cout << dp [0] << endl; } return 0; }
相关文章推荐
- ubuntu中mysql修改编码utf8
- android BitmapUtil 工具类
- ListView 滑动时数据重复错位解决
- 题目:在O(1)时间复杂度删除链表节点
- PHP RSA加解密示例(转)
- CentOS中安装subversion,并使用svn+ssh访问
- 题目:哈希函数
- 题目:合并排序数组 II
- HDU 4403 A very hard Aoshu problem (DFS)
- 主项定理Master Method
- 题目:合并排序数组
- Android播放gif动画,增加屏幕掉金币效果
- 使用C/C++发展Web系统开源
- Java String 与 StringBuffe 区别
- Android播放gif动画,增加屏幕掉金币效果
- 背包模板
- 【栈】日志分析(BSOJ2981)
- mysql中的主从复制slave-skip-errors参数使用方法
- 题目:合并区间
- MongoDB学习三--MongoDB简单增删改查