zoukankan      html  css  js  c++  java
  • CF 230A Dragons

    C - Dragons
    Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

    Description

    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 ≤ 1041 ≤ n ≤ 103). Then n lines follow: the i-th line contains space-separated integers xi and yi (1 ≤ xi ≤ 1040 ≤ 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.

    Sample Input

    Input
    2 2
    1 99
    100 0
    Output
    YES
    Input
    10 1
    100 100
    Output
    NO

    Hint

    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.

     1 #include <cstdio>
     2 #include <cstdlib>
     3 using namespace std;
     4 struct dragon
     5 {
     6   int x, y;
     7 }d[1000+10];
     8 int cmp(const void *a, const void *b)
     9 {
    10   return (*(dragon*)a).x > (*(dragon*)b).x ? 1 : -1;
    11 }
    12 int main(void){
    13   int s, n;
    14 #ifndef ONLINE_JUDGE
    15   freopen("dragon.in", "r", stdin);
    16 #endif
    17   while (~scanf("%d%d", &s, &n)){
    18     for (int i = 0; i < n; ++i){scanf("%d%d", &d[i].x, &d[i].y);}
    19     qsort(d, n, sizeof(d[0]), cmp);
    20     bool flag = true;
    21     for (int i = 0; i < n; ++i){
    22       if (s > d[i].x) s += d[i].y;
    23       else {flag = false; break;}
    24     }
    25     if (flag) printf("YES\n");
    26     else printf("NO\n");
    27   }
    28   return 0;
    29 }

    简单题,木有卡……

  • 相关阅读:
    tensorflow学习笔记----TensorBoard讲解
    tensorflow学习笔记----tensorflow在windows的安装及TensorBoard中mnist样例
    vmware workstation环境下虚拟机固定ip的设置
    一致性协议之Paxos算法
    一致性协议之二阶段提交和三阶段提交
    红黑树
    kibana增加验证
    Linux安装nodejs和npm
    Gnu pgp加密解密
    linux记录每次登陆的历史命令
  • 原文地址:https://www.cnblogs.com/liuxueyang/p/2952808.html
Copyright © 2011-2022 走看看