hdu2069Coin Change（母函数带限制总数量）

╰+哭是因爲堅強的太久メ 2022-07-30 20:21 180阅读 0赞

# Coin Change #

**Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 17093    Accepted Submission(s): 5849**

Problem Description

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, or two 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 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 100 coins.

Input

The input file contains any number of lines, each one consisting of a number ( ≤250 ) 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

题意：给你个硬币用1 5 10 25 50来兑换，兑换的总数量不能超出100；求有多少兑换方法。

\#include<iostream>  
\#include<cstdio>  
\#include<cstring>  
using namespace std;  
const int N=300;  
int c1\[N\]\[110\],c2\[N\]\[110\];  
int res\[N\],money\[6\]=\{0,1,5,10,25,50\};  
void Init()  
\{  
    memset(c1,0,sizeof(c1));  
    memset(c2,0,sizeof(c2));  
    c1\[0\]\[0\]=1;  
    for(int i=1;i<=5;i++)//提供1 5 10 25 50  
    \{  
        for(int j=0;j<=250;j++)//硬币最多达到250  
            for(int k=0;j+k\*money\[i\]<=250;k++)//提供下一个物品数量。  
                for(int p=0;k+p<=100;p++)       //限制硬币总数不超过100  
                    c2\[j+k\*money\[i\]\]\[p+k\]+=c1\[j\]\[p\];//dp\[i\]\[j\]代表价值为i兑换个数为j的兑换方式的个数。  
        for(int j=0;j<=250;j++)  
            for(int p=0;p<=100;p++)  
            \{    //限制硬币总数不超过100  
                c1\[j\]\[p\]=c2\[j\]\[p\];  
                c2\[j\]\[p\]=0;  
            \}  
    \}  
    for(int i=1;i<=250;i++)  
        for(int j=0;j<=100;j++)  
            res\[i\]+=c1\[i\]\[j\];  
    res\[0\]=1;       //别忘了这个  
\}  
  
  
int main()\{  
  
  
    //freopen("input.txt","r",stdin);  
    int n;  
    Init();  
    while(~scanf("%d",&n))  
    \{  
        printf("%d\\n",res\[n\]);  
    \}  
    return 0;  
\}