Description
llk and wy often go to the yh restaurant to have rice with toppings. One day after they finished eating, llk paid the two people's money together, but wy did not want to owe llk the money. Now wy has some loose money in his hands, and llk also has some loose money in his hands. wy wants to know if they can just make the two owe them, but wy is stupid, can you help wy?

Input
For multiple groups of test data, enter 3 non-negative integers, C, n, m in the first line of each group. C represents the money wy owes llk, n represents the type of face value of money in wy's hands, and m represents the type of face value of money in llk's hands. In the next n rows, each row contains two numbers v and c, which respectively represent c coins with face value v in wy's hand. In the next m rows, there are two numbers v and c in each row, representing c coins with face value v in the hand of llk. (C <= 10000; 1<=n, m<50; 0<=v <=100; 0<=c<=10)

Output
Each group of data outputs one line. If there is a scheme to make wy and llk not owe, output YES, otherwise output NO.

Sample Input
7 1 1
10 1
1 10
Sample Output
YES
HINT
wy gave llk a 10 yuan, and llk gave wy three 1 yuan

Idea: I wanted to use dfs at first, but it was too difficult to write. The discussion after the game was a multiple backpack. The test is dp judgment.

State definition: dp[i][j] represents the legal situation when the repayment amount of the first i category is j.

State transition: dp[i][j]=dp[i-1][j-k*v[i]] (k:0~c[i])

The usual dp is a little different, after the initial assignment of 1, and then dp[i][j] is 1, there is no need to transfer. And pay attention to the transfer conditions are j-k*v[i]>=0 (the amount of money to be paid back is positive) and j-k*v[i]<=C, my input does this to prevent crossing the boundary;

Then look at the code comments specifically, the school's oj is easy to Tle

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=150;
typedef int LL;//Int is required, otherwise T 
LL dp[maxn][10010];
LL v[maxn];
LL c[maxn];
int main(void)
{
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	LL C,n,m;
	while(cin>>C>>n>>m)
	{
		memset(dp,0,sizeof(dp));
		for(LL i=1;i<=n;i++)
			 cin>>v[i]>>c[i];
		for(LL i=n+1;i<=n+m;i++)
			{
				cin>>v[i]>>c[i];
				v[i]=-v[i];	
			}	 
		dp[0][0]=1;
		for(LL i=1;i<=n+m;i++)
			 for(LL j=0;j<=C;j++)//If j<=10000, change the loop of j and k, otherwise it will be T 
				for(LL k=0;k<=c[i];k++)/// 
				{
					 if(dp[i][j]) continue;//If it is already 1, don't update 
					 if(j-k*v[i]>=0&&j-k*v[i]<=C)//v[i] will be a negative number, <=C, otherwise it may exceed memory 
					dp[i][j]=dp[i-1][j-k*v[i]];
				}
		if(dp[n+m][C]) cout<<"YES"<<endl;
		else cout<<"NO"<<endl;	
	}
return 0;
}