【折半枚举】HDU_5616_Jam's balance
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Jam's balance
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1830 Accepted Submission(s): 767
Problem Description
Jim has a balance and N weights. (1≤N≤20)
The balance can only tell whether things on different side are the same weight.
Weights can be put on left side or right side arbitrarily.
Please tell whether the balance can measure an object of weight M.
The balance can only tell whether things on different side are the same weight.
Weights can be put on left side or right side arbitrarily.
Please tell whether the balance can measure an object of weight M.
Input
The first line is a integer T(1≤T≤5) , means T test cases.
For each test case :
The first line isN , means the number of weights.
The second line areN number, i'th number wi(1≤wi≤100) means the i'th weight's weight is wi .
The third line is a numberM . M is the weight of the object being measured.
For each test case :
The first line is
The second line are
The third line is a number
Output
You should output the "YES"or"NO".
Sample Input
121 43245
Sample Output
NOYESYESHintFor the Case 1:Put the 4 weight aloneFor the Case 2:Put the 4 weight and 1 weight on both side
Source
BestCoder Round #70
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hujie
#include <bits/stdc++.h>using namespace std;const int maxn=31;int a[maxn];set<int>s1,s2,temp;set<int>::iterator it;int main(){ int t,n,q,x; scanf("%d",&t); while(t--){ s1.clear();s2.clear(); s1.insert(0);s2.insert(0); scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d",&a[i]); for(int i=0;i<n/2;i++){ temp.clear(); for(it=s1.begin();it!=s1.end();it++){ temp.insert((*it)+a[i]); if((*it)-a[i]>=0) temp.insert((*it)-a[i]); else temp.insert(a[i]-(*it)); } for(it=temp.begin();it!=temp.end();it++) s1.insert((*it)); } for(int i=n/2;i<n;i++){ temp.clear(); for(it=s2.begin();it!=s2.end();it++){ temp.insert((*it)+a[i]); if((*it)-a[i]>=0) temp.insert((*it)-a[i]); else temp.insert(a[i]-(*it)); } for(it=temp.begin();it!=temp.end();it++) s2.insert((*it)); } scanf("%d",&q); while(q--){ scanf("%d",&x); if(x>2000){ puts("NO"); continue; } int f=0; for(int i=0;i<=x;i++){ if(s1.find(i)!=s1.end()&&s2.find(x-i)!=s2.end()){ f=1; break; } } if(f){ puts("YES"); continue; } for(int i=0;i<=2000;i++){ if((s1.find(i)!=s1.end()&&s2.find(i+x)!=s2.end()) ||(s2.find(i)!=s2.end()&&s1.find(i+x)!=s1.end())){ f=1; break; } } if(f) puts("YES"); else puts("NO"); } } return 0;}#include <bits/stdc++.h>using namespace std;const int maxn=31;int a[maxn];vector<int>v;bool vis[2010];int main(){ int t,n,q,x; scanf("%d",&t); while(t--){ v.clear(); memset(vis,false,sizeof(vis)); scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%d",&a[i]); a[n+i]=-a[i]; } v.push_back(0); for(int i=0;i<2*n;i++){ int t=v.size(); for(int j=0;j<t;j++){ int temp=v[j]+a[i]; if(temp>=0&&!vis[temp]){ vis[temp]=1; v.push_back(temp); } } } scanf("%d",&q); while(q--){ scanf("%d",&x); if(!vis[x]) printf("NO\n"); else printf("YES\n"); } }}
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