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Description
After making a purchase at a large department store, Mel‘s change was 17 cents. He received 1 dime, 1 nickel, and 2 pennies. Later that day, he was shopping at a convenience store. Again his change was 17 cents. This time he received 2 nickels and 7 pennies. He began to wonder ‘ "How many stores can I shop in and receive 17 cents change in a different configuration of coins? After a suitable mental struggle, he decided the answer was 6. He then challenged you to consider the general problem.
Write a program which will determine the number of different combinations of US coins (penny: 1c, nickel: 5c, dime: 10c, quarter: 25c, half-dollar: 50c) which may be used to produce a given amount of money.
The input will consist of a set of numbers between 0 and 30000 inclusive, one per line in the input file.
The output will consist of the appropriate statement from the selection below on a single line in the output file for each input value. The number m is the number your program computes, n is the input value.
There are mways to produce ncents change.
There is only 1 way to produce ncents change.
17 11 4
There are 6 ways to produce 17 cents change. There are 4 ways to produce 11 cents change. There is only 1 way to produce 4 cents change.
就是给你一个数，分别用1 5 10 25 50 相加有多少种方法可以加到这个数。
思路：
一个简单的完全背包问题，每种物品有无限件，还是根据01背包改变下次次序就可以。注意输出，这个是个坑。
当数为0的时候输出的也是1。
代码：
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <queue>
#include <map>
#include <cmath>
#include <string>
#define INF 0x3f3f3f3f
using namespace std;
int main()
{
int coin[5] = {1, 5, 10, 25, 50},i,j,n;
long long dp[30010];
while(scanf("%d",&n) != EOF)
{
for(i = 1; i <= n; i++)
dp[i] = 1;
dp[0] = 1;
for(i = 1; i < 5; i++)
{
for(j = coin[i]; j <= n; j++)
{
dp[j] = dp[j - coin[i]] + dp[j];
}
}
if(dp[n] != 1)
printf("There are %lld ways to produce %d cents change.\n",dp[n],n);
else
printf("There is only %lld way to produce %d cents change.\n",dp[n],n);
}
return 0;
}
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原文地址：http://blog.csdn.net/bmamb/article/details/51332810
