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 Description
Nowadays, 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).
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).
Input
Input 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.
A test case starting with a negative integer terminates input and this test case is not to be processed.
Output
For 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 should be as equal as possible. At the same time, you should guarantee that A is not less than B.
Sample Input
2
10 1
20 1
3
10 1
20 2
30 1
-1
Sample Output
20 10
40 40
Author
lcy
Mean:
有n种物品，第一种物品的单价为v1,数量为m1;第二种物品的单价为v2，数量为m2.....现在要你将这些物品分为两堆，使得这两堆物品的价值尽量接近，输出两堆物品的价值。
analyse:
这道题的解法很多：dp,母函数.....，详见《编程之美》。
这里我用了两种方法来做了一下，发现后面的方法比母函数快多了。
Time complexity：O(n^2)
Source code：
母函数代码：
// 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 long
using 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;
}
数学方法：
// 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 long
using 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;
}