HDU 6092 Rikka with Subset 【dp多重背包】【好题】
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Rikka with Subset
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 139 Accepted Submission(s): 49
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
题意:有一个数列 a[] ,长度(n<=50)。b[i] 表示元素和为 i 的集合个数。给你一个数列 b[] ,长度(m<=10000),让你求 a[],并按照其字典序最小输出
显然数字0的数量num[0]为log2(b[0]),数字1的数量num[1]为b[1]/b[0]
设dp[i],表示在当前i没有的情况下,用前面已知数量的数组成数字i共有多少种情况
那么b[i]-dp[i]即为数字i与0进行组合的可能性,则num[i]=(b[i]-dp[i])/b[0]
这里如果直接写的话复杂度为o(m^2)会超时,所以需要剪枝:
if (dp[j] == 0) continue;if (num[i] == 0) break;直接将m^2的复杂度降为nm
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <queue>#include <map>#define ms(a,b) memset(a,b,sizeof(a)) using namespace std;typedef long long ll;const int maxn = 1e4 + 100;ll b[maxn];int dp[maxn], num[maxn];int C(int n, int m){int sum = 1;for (int i = n - m + 1; i <= n; i++) sum *= i;for (int i = 1; i <= m; i++) sum /= i;return sum;}int main(){int t;scanf("%d", &t);while (t--){int n, m;scanf("%d%d", &n, &m);ms(dp, 0);ms(num, 0);for (int i = 0; i <= m; i++){scanf("%lld", &b[i]);}num[0] = log2(b[0]);num[1] = b[1] / b[0];dp[0] = b[0];for (int i = 0; i <= m; i++){for (int j = m; j >= 0; j--){if (dp[j] == 0) continue;if (num[i] == 0) break;for (int k = 1; k <= num[i]; k++){if (j + k*i <= m){dp[j + k*i] += dp[j] * C(num[i], k);}}}if (i + 1 <= m){num[i + 1] = (b[i + 1] - dp[i + 1]) / b[0];}}bool flag = 0;for (int i = 0; i <= m; i++){for (int j = 1; j <= num[i]; j++){if (!flag) printf("%d", i), flag = 1;else printf(" %d", i);}}puts("");}return 0;}
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