浙江中医药大学暑期训练测试赛八I
来源:互联网 发布:windows 10 恢复备份 编辑:程序博客网 时间:2024/05/16 18:07
1373: Crixalis's Equipment
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 138 Solved: 43
[Submit][Status][Web Board]
Description
Crixalis - Sand King used to be a giant scorpion(蝎子) in the deserts of Kalimdor. Though he's a guardian of Lich King now, he keeps the living habit of a scorpion like living underground and digging holes.
Someday Crixalis decides to move to another nice place and build a new house for himself (Actually it's just a new hole). As he collected a lot of equipment, he needs to dig a hole beside his new house to store them. This hole has a volume of V units, and Crixalis has N equipment, each of them needs Ai units of space. When dragging his equipment into the hole, Crixalis finds that he needs more space to ensure everything is placed well. Actually, the ith equipment needs Bi units of space during the moving. More precisely Crixalis can not move equipment into the hole unless there are Bi units of space left. After it moved in, the volume of the hole will decrease by Ai. Crixalis wonders if he can move all his equipment into the new hole and he turns to you for help.
Input
The first line contains an integer T, indicating the number of test cases. Then follows T cases, each one contains N + 1 lines. The first line contains 2 integers: V, volume of a hole and N, number of equipment respectively. The next N lines contain N pairs of integers: Ai and Bi.
0<T<= 10, 0<V<10000, 0<N<1000, 0 <Ai< V, Ai <= Bi < 1000.
Output
For each case output "Yes" if Crixalis can move all his equipment into the new hole or else output "No".
Sample Input
Sample Output
#include<algorithm>
#include<string.h>
#include<stdio.h>
#include<math.h>
using namespace std;
struct node
{
int x;
int y;
int c;
}a[1005];
bool cmp(node a,node b)
{
return a.c>b.c;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
a[i].c=a[i].y-a[i].x;
}
sort(a,a+m,cmp);
int flag=0;
for(int i=0;i<m;i++)
{
if(n>=a[i].y)
{
n-=a[i].x;
}
else
{
flag=1;
break;
}
}
if(flag)
printf("No\n");
else
printf("Yes\n");
}
return 0;
}
- 浙江中医药大学暑期训练测试赛八I
- 浙江中医药大学暑期训练测试赛八A
- 浙江中医药大学暑期训练测试赛八B
- 浙江中医药大学暑期训练测试赛八C
- 浙江中医药大学暑期训练测试赛八F
- 浙江中医药大学暑期训练测试赛八H
- 浙江中医药大学暑期训练测试赛八G
- 浙江中医药大学暑期训练测试赛八E
- 浙江中医药大学暑期训练测试赛十
- 浙江中医药大学暑期训练测试赛十 英雄无敌3(2) 数学公式
- 1591 浙江中医药大学
- 1531 浙江中医药大学
- 1066 浙江中医药大学ACM
- 1049 '最爱' 浙江中医药大学
- 1065 浙江中医药大学ACM OJ
- 寻找zcmu-2017浙江中医药大学程序设计
- 2017年浙江中医药大学大学生程序设计竞赛(重现赛)
- 2017年浙江中医药大学大学生程序设计竞赛(重现赛)-H剪纸
- ospf lsu报文格式
- 扩展欧几里得算法详解
- 怎么清理mac上的重复文件和重复的照片 Gemini 2
- 小白笔记--------------------leetcode27. Remove Element
- Nginx反向代理和负载均衡的一点想法以及常见错误
- 浙江中医药大学暑期训练测试赛八I
- spark和storm的对比
- android 前后台切换 回调
- 函数使用
- hadoop中namenode不能启动,导致浏览器不能连接
- JAVA——比较运算符
- Nginx反向代理设置从80端口转向其他端口
- 使用Myeclipse搭建的第一个SSH例子
- Python中打印如何不换行?