Codeforce 230A - Dragons （sort）

Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all n dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals s.

If Kirito starts duelling with the i-th (1 ≤ i ≤ n) dragon and Kirito's strength is not greater than the dragon's strength xi, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by yi.

Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.

Input

The first line contains two space-separated integers s and n (1 ≤ s ≤ 104, 1 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 104, 0 ≤ yi ≤ 104) — the i-th dragon's strength and the bonus for defeating it.

Output

On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.

Examples
input
2 2
1 99
100 0
output
YES
input
10 1
100 100
output
NO
Note

In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.

In the second sample Kirito's strength is too small to defeat the only dragon and win.

 题解：不多说 水水

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <vector>
 6 #include <cstdlib>
 7 #include <iomanip>
 8 #include <cmath>
 9 #include <ctime>
10 #include <map>
11 #include <set>
12 using namespace std;
13 #define lowbit(x) (x&(-x))
14 #define max(x,y) (x>y?x:y)
15 #define min(x,y) (x<y?x:y)
16 #define MAX 100000000000000000
17 #define MOD 1000000007
18 #define pi acos(-1.0)
19 #define ei exp(1)
20 #define PI 3.141592653589793238462
21 #define INF 0x3f3f3f3f3f
22 #define mem(a) (memset(a,0,sizeof(a)))
23 typedef long long ll;
24 const int N=1005;
25 const int mod=1e9+7;
26 struct Node
27 {
28     int x, y;
29 }a[N];
30 bool cmp(Node i,Node j)
31 {
32     return i.x<j.x;
33 }
34 int main()
35 {
36     int s,n;
37     scanf("%d %d",&s,&n);
38     for(int i=0;i<n;i++){
39         scanf("%d %d",&a[i].x,&a[i].y);
40     }
41     sort(a,a+n,cmp);
42     int flag=1;
43     for(int i=0;i<n;i++){
44         if(a[i].x>=s){
45             flag=0;
46             break;
47         }
48         else{
49             s+=a[i].y;
50         }
51     }
52     if(flag) cout<<"YES"<<endl;
53     else cout<<"NO"<<endl;
54     return 0;
55 }
View Code

 

posted @ 2017-07-30 11:33  wydxry  阅读(433)  评论(0)  编辑  收藏  举报
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