hud Rikka with Subset 2017 多校第五场 (背包)
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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 is 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).
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.
Sample Input
2
2 3
1 1 1 1
3 3
1 3 3 1
Sample Output
1 2
1 1 1
Hint
In the first sample,
题目大意:给你一个a数组,a数组中所有元素之和是m,b【i】表示0—m中a数组的子集和为i的总次数。如第一组测试样例,a数组的和为3且只有两个元素,所以a={1,2}它的子集有空集,{1},{2},{1,2},a的子集中所有元素之分别和为0,1,2,3,所以0,1,2,3出现的次数是1,所以b数组全为1.
解题思路:如果 B[i]是 B数组中除了 B[0]以外第一个值不为 0 的位置,那么显然 i 就是 A 中的最小数。现在需要求出删掉 i后的 B 数组,过程大概是反向的背包,即从小到大让 B[j]-=B[j-i].
时间复杂度 O(nm).
#include <bits/stdc++.h>typedef long long ll;using namespace std;ll b[10005];int a[55];int main(){ int T;scanf("%d",&T); while(T--){ int n,m,k; scanf("%d %d",&n,&m); for(int i=0;i<=m;i++){ scanf("%lld",&b[i]); } for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) if(b[j]!=0) { k = j;break; } a[i] = k; for(int j = k;j<=m;j++) b[j] -= b[j-k]; } printf("%d",a[1]); for(int i=2;i<=n;i++) printf(" %d",a[i]); printf("\n"); } return 0;}
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