Jobdu 1209最小邮票数 - KingEasternSun的专栏 - 博客频道 - CSDN.NET


九度笔记之 1209最小邮票数 - KingEasternSun的专栏 - 博客频道 - CSDN.NET
    有若干张邮票,要求从中选取最少的邮票张数凑成一个给定的总值。
    如,有1分,3分,3分,3分,4分五张邮票,要求凑成10分,则使用3张邮票:3分、3分、4分即可。
输入:
    有多组数据,对于每组数据,首先是要求凑成的邮票总值M,M<100。然后是一个数N,N〈20,表示有N张邮票。接下来是N个正整数,分别表示这N张邮票的面值,且以升序排列。
输出:
      对于每组数据,能够凑成总值M的最少邮票张数。若无解,输出0。
样例输入:
10 5 1 3 3 3 4
样例输出:
3
void minNum(int m,int n){
    int *dp = new int[m+1];
    dp[0] = 0;
    for(int i = 1;i<=m;i++)
        dp[i] = INF;
    int *num = new int[n];
    for(int i = 0;i<n;i++)
        std::cin>>num[i];
    for(int i = 0;i<n;i++){
        for(int j = m;j>=num[i];j--){ //must from m to num[i]
            dp[j] = std::min(dp[j],dp[j-num[i]]+1);
        }
    }
    if(dp[m]<INF)
        std::cout<<dp[m]<<std::endl;
    else
        std::cout<<0<<std::endl;
}
void minNumnew(int m,int n){
    int *dp = new int[m+1];
    dp[0] = 0;
    for(int i = 1;i<=m;i++)
        dp[i] = INF;
 
    int num;
    for(int i = 0;i<n;i++){
        std::cin>>num;
        for(int j = m;j>=num;j--){ //must from m to num[i]
            dp[j] = std::min(dp[j],dp[j-num]+1);
        }
    }
    if(dp[m]<INF)
        std::cout<<dp[m]<<std::endl;
    else
        std::cout<<0<<std::endl;
}
http://www.acmerblog.com/jiudu-1209-2365.html
10    public static void main(String[] args) {
11        Scanner s = new Scanner(new BufferedInputStream(System.in));
12        while(s.hasNextInt()){
13            sum = s.nextInt();
14            n = s.nextInt();
15            arr = new int[n];
16            opt = new int[n+1][sum+1];
17            for(int i=0; i<n; i++)
18                arr[i] = s.nextInt();
19            Arrays.sort(arr);
20            int r = f(n-1,sum);
21            if(r >= MAX)
22                System.out.println(0);
23            else
24                System.out.println(r);
25        }
26    }
27    static int f(int i,int sum){
28        if(sum <= 0)
29            return MAX;
30        if(opt[i][sum] > 0)
31            return opt[i][sum];
32        if(Arrays.binarySearch(arr, 0, i+1, sum) >= 0){
33            opt[i][sum] = 1;
34            return opt[i][sum];
35        }
36        else if(i > 0){
37            return opt[i][sum]=Math.min(f(i-1,sum-arr[i])+1, f(i-1,sum));
38        }
39        if(i==0)
40            return opt[i][sum]= (arr[0] == sum ? 1:MAX);
41        return opt[i][sum];
42    }
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