1068 Find More Coins (30 分)
标签：获得 要求 stream NPU 回顾 wrap round 意义 line
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.
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 Nface values of the coins, which are all positive numbers. All the numbers in a line are separated by a space.
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].
8 9
5 9 8 7 2 3 4 1
1 3 5
4 8
7 2 4 3
No Solution
题意：
用n个硬币买价值为m的东西，输出使用方案，使得正好几个硬币加起来价值为m。从小到大排列，输出最小的那个排列方案
题解：
这是一道01背包问题，解题时注意题意的转化：
AC代码:
#include<iostream> #include<queue> #include<algorithm> using namespace std; int dp[101];//bags[i]面值不大于i的最大的面值和 bool choice[10001][101]; int cmp(int a, int b){return a > b;} int main(){ int w[10001]; int n,m; cin>>n>>m; for(int i=1;i<=n;i++){ cin>>w[i]; } sort(w+1,w+1+n,cmp); for(int i=1;i<=n;i++){ for(int j=m;j>=w[i];j--){ //之所以要反着来，和背包问题的更新规则有关 if(dp[j-w[i]]+w[i]>=dp[j]){//等号必须取到，否则输出的解是最大的sequence choice[i][j]=true;//跟踪哪个物品被选择了 dp[j]=dp[j-w[i]]+w[i]; } } } if(dp[m] != m) printf("No Solution"); else{ //下面是输出最优组合的过程，其实和背包问题的更新规则有关，就是沿着选出解的路径，反着走回去，就找到了所有被选择的数字。 int i=n,j=m; while(1){ if(choice[i][j]==true){ cout<<w[i]; j-=w[i]; if(j!=0) cout<<" "; } i--; if(j==0||i==0) break; } } return 0; }
PAT 甲级 1068 Find More Coins (30 分) (dp，01背包问题记录最佳选择方案)***
标签：获得 要求 stream NPU 回顾 wrap round 意义 line
原文地址：https://www.cnblogs.com/caiyishuai/p/11762267.html
