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"Fat and docile, big and dumb, they look so stupid, they aren‘t much
fun..."
- Cows with Guns by Dana Lyons

The cows want to prove to the public that they are both smart and fun. In order to do this, Bessie has organized an exhibition that will be put on by the cows. She has given each of the N (1 <= N <= 100) cows a thorough interview and determined two values for each cow: the smartness Si (-1000 <= Si <= 1000) of the cow and the funness Fi (-1000 <= Fi <= 1000) of the cow.

Bessie must choose which cows she wants to bring to her exhibition. She believes that the total smartness TS of the group is the sum of the Si‘s and, likewise, the total funness TF of the group is the sum of the Fi‘s. Bessie wants to maximize the sum of TS and TF, but she also wants both of these values to be non-negative (since she must also show that the cows are well-rounded; a negative TS or TF would ruin this). Help Bessie maximize the sum of TS and TF without letting either of these values become negative.

Input

* Line 1: A single integer N, the number of cows

* Lines 2..N+1: Two space-separated integers Si and Fi, respectively the smartness and funness for each cow.

Output

* Line 1: One integer: the optimal sum of TS and TF such that both TS and TF are non-negative. If no subset of the cows has non-negative TS and non- negative TF, print 0.

Sample Input

5
-5 7
8 -6
6 -3
2 1
-8 -5

Sample Output

8

Hint

OUTPUT DETAILS:

Bessie chooses cows 1, 3, and 4, giving values of TS = -5+6+2 = 3 and TF
= 7-3+1 = 5, so 3+5 = 8. Note that adding cow 2 would improve the value
of TS+TF to 10, but the new value of TF would be negative, so it is not
allowed. 題目大意： 給定n頭牛的TS值和TF值，求最大的TS+TF值，前提是TS的和與TF的和不為負數。 由於有負數，所以整體移動1e5，即1e5相當於0。

#include <iostream>
#include <cstring>
using namespace std;
const int INF=0x3f3f3f3f;
const int mid=1e5;
int dp[200005],s[105],f[105];
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>s[i]>>f[i];
    fill(dp,dp+200005,-INF);
    dp[mid]=0;
    for(int i=1;i<=n;i++)
    {
        if(s[i]>0)
        {
            for(int j=200004;j>=s[i];j--)
                dp[j]=max(dp[j],dp[j-s[i]]+f[i]);
        }
        else
        {
            for(int j=0;j-s[i]<200005;j++) ///j-s[i]>j因此遞增
                dp[j]=max(dp[j],dp[j-s[i]]+f[i]);
        }
    }
    int ans=0;
    for(int i=mid;i<200005;i++)
        if(dp[i]>=0)
            ans=max(ans,dp[i]+i-mid);
    cout<<ans<<‘\n‘;
    return 0;
}

Cow Exhibition （01背包）