HDU6092-Rikka with Subset
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Rikka with Subset
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1535 Accepted Submission(s): 775
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 hasn positive A1−An and their sum is m . Then for each subset S of A , Yuta calculates the sum of S .
Now, Yuta has got2n numbers between [0,m] . For each i∈[0,m] , he counts the number of i s he got as Bi .
Yuta shows Rikka the arrayBi and he wants Rikka to restore A1−An .
It is too difficult for Rikka. Can you help her?
Yuta has
Now, Yuta has got
Yuta shows Rikka the array
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 numbersn,m(1≤n≤50,1≤m≤104) .
The second line containsm+1 numbers B0−Bm(0≤Bi≤2n) .
For each testcase, the first line contains two numbers
The second line contains
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
22 31 1 1 13 31 3 3 1
Sample Output
1 21 1 1HintIn the first sample, $A$ is $[1,2]$. $A$ has four subsets $[],[1],[2],[1,2]$ and the sums of each subset are $0,1,2,3$. So $B=[1,1,1,1]$
Source
2017 Multi-University Training Contest - Team 5
解题思路:首先题目中0的个数和1的个数必然是确定的,根据1的个数,我们可以算出它对后面数字的贡献,然后后面这些数字减去这个贡献,到最后这些数字就是自己对自己的贡献,即每种数字有几个
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <unordered_map>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, m;LL c[55][55], b[10011];void C(){ int i, j; c[0][0] = 1; for(i = 1; i < 55; i++) { c[i][0] = c[i][i] = 1; for(j = 1; j < i; j++) { c[i][j] = c[i - 1][j - 1] + c[i - 1][j]; } }}int main(){ C(); int t; scanf("%d", &t); while(t--) { scanf("%d%d",&n,&m); for(int i = 0; i <= m; i++) scanf("%lld", &b[i]); for(int i = 1; i <= m; i++) { if(b[i] == 0) continue; for(int j = i + 1; j <= m; j++) { for(int k = 1; k <= b[i]; k++) { if(j - i * k < i) break; if(j - i * k != i) b[j] -= c[b[i]][k] * b[j - i * k]; else b[j] -= c[b[i]][k + 1]; } } } int flag=1; for(int i = 1; i <= m; i++) { while(b[i]--) { if(flag) flag=0; else printf(" "); printf("%d", i); } } printf("\n"); } return 0;}
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