题目描述
In a 10-dollar shop, everything is worthy 10 dollars or less. In order to serve customers more effectively at the cashier, change needs to be provided in the minimum number of coins.
In this problem, you are going to provide a given value of the change in different coins. Write a program to calculate the number of coins needed for each type of coin.
The input includes a value v, a size of the coinage set n, and a face value of each coin, f1, f2, …, fn. The output is a list of numbers, namely, c1, …, cn, indicating the number of coins needed for each type of coin. There may be many ways for the change. The value v is an integer satisfying 0 < v ≤ 2000, representing the change required
in cents. The face value of a coin is less than or equal to 10000. The output of your program should take the combination with the least number of coins needed.
For example, the Hong Kong coinage issued by the Hong Kong Monetary Authority consists of 10 cents, 20 cents, 50 cents, 1 dollar, 2 dollars, 5 dollars and 10 dollars would be represented in the input by n = 7, f1 = 10, f2 = 20, f3 = 50, f4 = 100, f5 = 200, f6 = 500, f7 = 1000.
输入
The test data may contain many test cases, please process it to the end of the file.
Each test case contains integers v, n, f1, …, fn in a line. It is guaranteed that n ≤ 10 and 0 < f1 < f2 < …< fn.
输出
The output be n numbers in a line, separated by space. If there is no possible change, your output should be a single −1. If there are more than one possible solutions, your program should output the one that uses more coins of a lower face value.
样例输入
2000 7 10 20 50 100 200 500 1000
250 4 10 20 125 150
35 4 10 20 125 150
48 4 1 8 16 20
40 4 1 10 13 37
43 5 1 2 21 40 80
样例输出
0 0 0 0 0 0 2
0 0 2 0
-1
0 1 0 2
3 0 0 1
1 1 0 1 0
题目大意：
给出一个钱的总数以及硬币的面额数，每种硬币的使用次数并没有限制，在保证能够用硬币得到总的金额的情况下，输出最小使用硬币的情况下每种硬币的使用个数，如果不能用所给出的硬币得到总的金额就输出 -1
对于前面判断能不能用所给面值的硬币得到总的金额用完全背包就很容易就可以解决，重点是如何进行路径的记录
int n, sum, a[maxn];
int ans[maxn], dp[maxn], cur[maxn];
int main()
{
while (cin >> sum >> n) {
memset(dp, 0x3f3f3f3f, sizeof(dp));
memset(ans, 0, sizeof(ans));
for (int i = 1; i <= n; i++) a[i] = read;
dp[0] = 0;
for (int i = 1; i <= n; i++) {
for (int j = a[i]; j <= sum; j++) {
if (dp[j - a[i]] < INF && dp[j - a[i]] < dp[j]) {
dp[j] = dp[j - a[i]] + 1;
cur[j] = j - a[i];///存下当前的
}
}
}
///cout<<dp[sum]<<endl;
if (dp[sum] == 0x3f3f3f3f) puts("-1");
else {
ll t = sum;///记录总数
while (t) {
ans[t - cur[t]] ++;///每一个的数量
t = cur[t];
}
for (int i = 1; i <= n; i++) {
printf("%lld", ans[a[i]]);
if (i != n) printf(" ");
}
puts("");
}
}
return 0;
}