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组合数学 - 母函数的运用 --- hdu 1709 :The Balance
The Balance
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5706 Accepted Submission(s): 2311
Problem Description
Now you are asked to measure a dose of medicine with a balance and a number of weights. Certainly it is not always achievable. So you should find out the qualities which cannot be measured from the range [1,S]. S is the total quality of all the weights.
Input
The input consists of multiple test cases, and each case begins with a single positive integer N (1<=N<=100) on a line by itself indicating the number of weights you have. Followed by N integers Ai (1<=i<=N), indicating the quality of each weight where 1<=Ai<=100.
Output
For each input set, you should first print a line specifying the number of qualities which cannot be measured. Then print another line which consists all the irrealizable qualities if the number is not zero.
Sample Input
3
1 2 4
3
9 2 1
Sample Output
0
2
4 5
Mean:
给你N个砝码,每个砝码有自己的重量,现在要你计算从1~S中哪些重量是不能用这些砝码称出来的。(其中S为所有砝码的重量之和)。
analyse:
就是一道母函数的运用题,要注意的是砝码可以摆放在两个托盘上,所以要注意的是两个托盘两边的差值也能称出来的情况,其他的和普通母函数差不多。
Time complexity:O(n^3)
Source code:
// Memory Time// 1347K 0MS// by : Snarl_jsb// 2014-09-19-11.59#include<algorithm>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<vector>#include<queue>#include<stack>#include<map>#include<string>#include<climits>#include<cmath>#define N 10100#define LL long longusing namespace std;int val[N];int c1[N],c2[N];int main(){ ios_base::sync_with_stdio(false); cin.tie(0);// freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);// freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout); int n; while(cin>>n) { long long sum=0; for(int i=1;i<=n;++i) { cin>>val[i]; sum+=val[i]; } memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2)); c1[0]=c1[val[1]]=1; for(int i=2;i<=n;++i) { for(int j=0;j<=sum;++j) { for(int k=0;k<=1;++k) { c2[k*val[i]+j]+=c1[j]; c2[abs(k*val[i]-j)]+=c1[j]; } } for(int j=0;j<=sum;++j) { c1[j]=c2[j]; c2[j]=0; } } int res[N],ans=0; for(int i=1;i<=sum;++i) { if(!c1[i]) { res[ans++]=i; } } if(!ans) { puts("0"); continue; } cout<<ans<<endl; ans--; for(int i=0;i<ans;++i) { cout<<res[i]<<" "; } cout<<res[ans]<<endl; } return 0;}
组合数学 - 母函数的运用 --- hdu 1709 :The Balance
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