Big Event in HDU
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).
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.
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
解题思路：
母函数，趁热打铁。 计算所有可能的取值的方案数。 那么方案大于0 就说明可以取到咯。 然后从 sum/2 开始找，找到一个能取到的就是最接近的值了。
AC代码：
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std
;
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n,m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n,m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sc(n) scanf("%c",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,a,n) for(int i=n;i>=a;i--)
#define mem(a,n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define mod(x) ((x)%MOD)
#define gcd(a,b) __gcd(a,b)
#define lowbit(x) (x&-x)
#define pii map<int,int>
#define mk make_pair
#define rtl rt<<1
#define rtr rt<<1|1
#define Max(x,y) (x)>(y)?(x):(y)
#define int long long
typedef pair
<int,int> PII
;
typedef long long ll
;
typedef unsigned long long ull
;
typedef long double ld
;
const int MOD
= 1e9 + 7;
const ll mod
= 10007;
const double eps
= 1e-9;
const ll INF
= 0x3f3f3f3f3f3f3f3fll;
inline int read(){int ret
= 0, sgn
= 1;char ch
= getchar();
while(ch
< '0' || ch
> '9'){if(ch
== '-')sgn
= -1;ch
= getchar();}
while (ch
>= '0' && ch
<= '9'){ret
= ret
*10 + ch
- '0';ch
= getchar();}
return ret
*sgn
;}
inline void Out(int a
){if(a
>9) Out(a
/10);putchar(a
%10+'0');}
int qpow(int m
, int k
, int mod
){int res
=1%mod
,t
=m
%mod
;while(k
){if(k
&1)res
=res
*t
%mod
;t
=t
*t
%mod
;k
>>=1;}return res
;}
ll
gcd(ll a
,ll b
){if(b
> a
) swap(a
,b
); return b
==0?a
: gcd(b
,a
%b
);}
ll
lcm(ll a
,ll b
){return a
/gcd(a
,b
)*b
;}
ll
inv(ll x
,ll mod
){return qpow(x
,mod
-2,mod
)%mod
;}
int t
= 1,cas
= 1;
int n
,m
;
const int N
= 5e5+3;
int a
[N
],b
[N
];
int c1
[N
],c2
[N
];
signed main()
{
while(cin
>>n
&& n
> 0)
{
int sum
= 0;
for(int i
= 0 ; i
< n
; i
++){
cin
>>a
[i
]>>b
[i
];
sum
+= a
[i
]*b
[i
];
}
for(int i
= 0 ; i
<= sum
; i
++){
c1
[i
] = c2
[i
] = 0;
}
for(int i
= 0 ; i
<= b
[0] ; i
++){
c1
[i
*a
[0]] = 1;
}
int maxx
= b
[0]*a
[0];
for(int i
= 1 ; i
< n
; i
++)
{
for(int j
= 0 ; j
<= maxx
; j
++){
for(int k
= 0 ; k
<= b
[i
] ; k
++){
c2
[j
+k
*a
[i
]] = c1
[j
];
}
}
maxx
+= b
[i
]*a
[i
];
for(int j
= 1; j
<= maxx
; j
++){
c1
[j
] = c2
[j
];
c2
[j
] = 0;
}
}
for(int i
= sum
/2 ; i
>= 0 ; i
--){
if(c1
[i
]){
cout
<<sum
-i
<<" "<<i
<<endl
;
break;
}
}
}
}
