PAT-A1068 Find More Coins 题目内容及题解

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she must pay the exact amount. Since she has as many as 10^4 coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find some coins to pay for it.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (≤10^4, the total number of coins) and M (≤10^2, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the face values V1≤V2≤⋯≤Vk such that V1+V2+⋯+Vk=M. All the numbers must be separated by a space, and there must be no extra space at the end of the line. If such a solution is not unique, output the smallest sequence. If there is no solution, output "No Solution" instead.

Note: sequence {A[1], A[2], ...} is said to be "smaller" than sequence {B[1], B[2], ...} if there exists k≥1 such that A[i]=B[i] for all i<k, and A[k] < B[k].

Sample Input 1:

8 9
5 9 8 7 2 3 4 1

Sample Output 1:

1 3 5

Sample Input 2:

4 8
7 2 4 3

Sample Output 2:

No Solution

题目大意

题目给定一组数据作为硬币的面值，要求对于某个给定值输出能否使用硬币来恰好支付，如能，则输出其面值，不如能，输出无解。

解题思路

读入并储存所有硬币面值，将其按照从大到小的顺序进行排序；
用动态规划（背包问题）的方式获取是否可能取得给定值；
如不能，输出无解；
如能，按照题目格式输出结果；
返回零值。

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int dp[10010], w[10010];
bool choice[10010][110];
bool cmp(int a,int b){
    return a>b;
}
int main() {
    int N,M;
    int i,j;
    int v,index;
    vector<int> pick;
    scanf("%d%d",&N,&M);
    for(i=0;i<N;i++){
        scanf("%d", &w[i]);
    }
    sort(w,w+N,cmp);
    for(i=0;i<N;i++) {
        for(j=M;j>=w[i];j--) {
            if(dp[j]<= dp[j-w[i]]+w[i]){
                choice[i][j]=1;
                dp[j]=dp[j-w[i]]+w[i];
            }
        }
    }
    if(dp[M]!=M){
        printf("No Solution\n");
    }else{
        v=M,index=N;
        while(v>0){
            if(choice[index][v]==1){
                pick.push_back(w[index]);
                v-=w[index];
            }
            index--;
        }
        for(i=0;i<pick.size();i++){
            if(i!=0){
                printf(" ");
            }
            printf("%d",pick[i]);
        }
        printf("\n");
    }
    return 0;
}

运行结果

PAT-A1068 Find More Coins 题目内容及题解