1068 Find More Coins （01背包+选择物品的输出）

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 1 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 (≤, the total number of coins) and M (≤, 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 V​1​​≤V​2​​≤⋯≤V​k​​ such that V​1​​+V​2​​+⋯+V​k​​=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

　　

一维形式的状态转移方程：

01背包问题逆序，完全背包正序

此题还要输出选择的序列，故需要一个二维数组choose[maxv][pay]来保存信息，想象单元格1为阶梯形

#include<cstdio>
#include<algorithm>
using namespace std;
const int numv=10010;
const int pay=110;
int dp[pay]={0},c[numv];
int choose[numv][pay];

bool cmp(int a,int b){
    return a>b;
}

int main(){
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%d",&c[i]);
    }
    sort(c+1,c+1+n,cmp);
    for(int i=1;i<=n;i++){
        for(int j=m;j>=c[i];j--){
            if(dp[j]<=dp[j-c[i]]+c[i]){//要使元素尽量小，等于也要放 
                dp[j]=dp[j-c[i]]+c[i];
                choose[i][j]=1;
            }
            else choose[i][j]=0;
        }
    }
    if(dp[m]!=m) printf("No Solution");
    else{
        int now=m,i=n;
        while(now>0){
            if(choose[i][now]==1){
                now-=c[i];
                if(now==0) printf("%d",c[i]);
                else printf("%d ",c[i]);
            }
            i--;
        }
    }
    return 0;
}