Combined mathematics-deformation of the primary function --- HDU 1171: big event in HDU

Big event in HDU

Time Limit: 10000/5000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 24002 accepted submission (s): 8458

Problem descriptionnowadays, we all know that computer College is the biggest department in HDU. But, maybe you don't know that computer College had ever been split into computer College and software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. all facilities must go halves. first, all facilities are assessed, and two facilities are thought to be same if they have the same value. it is assumed that there is n (0 <n <1000) kinds of facilities (different value, different kinds ).

 

Inputinput contains multiple test cases. each test case starts with a number N (0 <n <= 50 -- the total number of different facilities ). the next n lines contain an integer v (0 <v <= 50 -- Value of facility) and an integer m (0 <m <= 100 -- corresponding number of the facilities) each. you can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

 

Outputfor each case, print one line containing two integers A and B which denote the value of computer College and software college will get respectively. A and B shoshould be as equal as possible. at the same time, You shoshould guarantee that A is not less than B.

 

Sample input210 120 1310 1 20 230 1-1

 

Sample output20 1040 40 authorlcy

 

Mean: 

There are n items. The unit price of the first item is V1, and the quantity is M1. The unit price of the second item is V2, and the quantity is m2 ..... now you need to divide these items into two heaps, so that the value of these two heaps of items is as close as possible, and the value of these two heaps of items is output.

Analyze:

There are many solutions to this problem: DP, the primary function.... For details, see the beauty of programming.

Here I used two methods to do it, and found that the following method is much faster than the master function.

Time Complexity:O (N ^ 2)

 

Source code:

Code of the primary function:

// Memory   Time// 1347K     0MS// by : Snarl_jsb// 2014-09-18-18.50#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 234567#define LL long longusing namespace std;int val[600],cnt[110];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&&n>0)    {        long long sum=0;        for(int i=1;i<=n;++i)        {            cin>>val[i]>>cnt[i];            sum+=val[i]*cnt[i];        }        memset(c1,0,sizeof(c1));        memset(c2,0,sizeof(c2));        for(int i=0;i<=cnt[1]*val[1];i+=val[1])            c1[i]=1;        for(int i=2;i<=n;++i)        {            for(int j=0;j<=sum;++j)            {                for(int k=0;k<=cnt[i];++k)                {                    c2[val[i]*k+j]+=c1[j];                }            }            for(int j=0;j<=sum;++j)            {                c1[j]=c2[j];                c2[j]=0;            }        }        if(c1[sum/2]==2)        {            cout<<sum/2<<" "<<sum/2<<endl;            continue;        }         int t=sum/2;         int QAQ,TAT;         int minn=987654321;         for(int i=0;i<=sum;++i)         {             if(c1[i])             {                 if(abs(sum/2-i)<minn)                 {                     minn=abs(sum/2-i);                     QAQ=i;                 }             }         }         TAT=sum-QAQ;         if(QAQ<TAT)         {             QAQ^=TAT^=QAQ^=TAT;         }        cout<<QAQ<<" "<<TAT<<endl;    }    return 0;}

　　

Mathematical Methods:

// Memory   Time// 1347K     0MS// by : Snarl_jsb// 2014-09-18-23.05#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 1000010#define LL long longusing namespace std;int val[N],cnt[N];int buff[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&&n>0){long long v,t,idx=0,sum=0;for(int i=1;i<=n;i++){cin>>v>>t;val[i]=v,cnt[i]=t;sum+=val[i]*cnt[i];while(t--){buff[++idx]=v;}}sort(buff+1,buff+1+idx);int half=sum/2;long long ans=0;for(int i=idx;i>=1;--i){if(ans+buff[i]<=half){ans+=buff[i];}}cout<<sum-ans<<" "<<ans<<endl;}    return 0;}

　　

Combined mathematics-deformation of the primary function --- HDU 1171: big event in HDU