首页 > 代码库 > hdu6092 01背包
hdu6092 01背包
Rikka with Subset
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 658 Accepted Submission(s): 297
Problem Description
As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:
Yuta has n positive A1?An and their sum is m . Then for each subset S of A , Yuta calculates the sum of S .
Now, Yuta has got 2n numbers between [0,m] . For each i∈[0,m] , he counts the number of i s he got as Bi .
Yuta shows Rikka the array Bi and he wants Rikka to restore A1?An .
It is too difficult for Rikka. Can you help her?
Yuta has n positive A1?An and their sum is m . Then for each subset S of A , Yuta calculates the sum of S .
Now, Yuta has got 2n numbers between [0,m] . For each i∈[0,m] , he counts the number of i s he got as Bi .
Yuta shows Rikka the array Bi and he wants Rikka to restore A1?An .
It is too difficult for Rikka. Can you help her?
Input
The first line contains a number t(1≤t≤70)
, the number of the testcases.
For each testcase, the first line contains two numbers n,m(1≤n≤50,1≤m≤104) .
The second line contains m+1 numbers B0?Bm(0≤Bi≤2n) .
For each testcase, the first line contains two numbers n,m(1≤n≤50,1≤m≤104) .
The second line contains m+1 numbers B0?Bm(0≤Bi≤2n) .
Output
For each testcase, print a single line with n
numbers A1?An
.
It is guaranteed that there exists at least one solution. And if there are different solutions, print the lexicographic minimum one.
It is guaranteed that there exists at least one solution. And if there are different solutions, print the lexicographic minimum one.
Sample Input
2
2 3
1 1 1 1
3 3
1 3 3 1
Sample Output
1 2
1 1 1
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N=1e4+88; typedef long long LL; LL dp[N],B[N]; int a[N]; int main(){ int n,m,T; for(scanf("%d",&T);T--;){ memset(dp,0,sizeof(dp)); memset(a,0,sizeof(a)); scanf("%d%d",&n,&m); for(int i=0;i<=m;++i) scanf("%I64d",&B[i]); dp[0]=1; for(int i=1;i<=m;++i){ if(dp[i]==B[i]) continue; a[i]=B[i]-dp[i]; for(int j=1;j<=a[i];++j) for(int k=m;k>=i;--k) dp[k]+=dp[k-i]; } int i; for(i=1;i<=m;++i) if(a[i]--) {printf("%d",i);break;} for(;i<=m;++i) while(a[i]--) printf(" %d",i); puts(""); } }
hdu6092 01背包
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。