cf Educational Codeforces Round 85 (Rated for Div. 2)A. Level Statistics

2024-02-04 21:48

本文主要是介绍cf Educational Codeforces Round 85 (Rated for Div. 2)A. Level Statistics,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!

题目链接:https://codeforces.com/contest/1334/problem/A
A. Level Statistics
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Polycarp has recently created a new level in this cool new game Berlio Maker 85 and uploaded it online. Now players from all over the world can try his level.

All levels in this game have two stats to them: the number of plays and the number of clears. So when a player attempts the level, the number of plays increases by 1. If he manages to finish the level successfully then the number of clears increases by 1 as well. Note that both of the statistics update at the same time (so if the player finishes the level successfully then the number of plays will increase at the same time as the number of clears).

Polycarp is very excited about his level, so he keeps peeking at the stats to know how hard his level turns out to be.

So he peeked at the stats n times and wrote down n pairs of integers — (p1,c1),(p2,c2),…,(pn,cn), where pi is the number of plays at the i-th moment of time and ci is the number of clears at the same moment of time. The stats are given in chronological order (i.e. the order of given pairs is exactly the same as Polycarp has written down).

Between two consecutive moments of time Polycarp peeked at the stats many players (but possibly zero) could attempt the level.

Finally, Polycarp wonders if he hasn’t messed up any records and all the pairs are correct. If there could exist such a sequence of plays (and clears, respectively) that the stats were exactly as Polycarp has written down, then he considers his records correct.

Help him to check the correctness of his records.

For your convenience you have to answer multiple independent test cases.

Input
The first line contains a single integer T (1≤T≤500) — the number of test cases.

The first line of each test case contains a single integer n (1≤n≤100) — the number of moments of time Polycarp peeked at the stats.

Each of the next n lines contains two integers pi and ci (0≤pi,ci≤1000) — the number of plays and the number of clears of the level at the i-th moment of time.

Note that the stats are given in chronological order.

Output
For each test case print a single line.

If there could exist such a sequence of plays (and clears, respectively) that the stats were exactly as Polycarp has written down, then print “YES”.

Otherwise, print “NO”.

You can print each letter in any case (upper or lower).

Example

input
6
3
0 0
1 1
1 2
2
1 0
1000 3
4
10 1
15 2
10 2
15 2
1
765 432
2
4 4
4 3
5
0 0
1 0
1 0
1 0
1 0
output
NO
YES
NO
YES
NO
YES
Note
In the first test case at the third moment of time the number of clears increased but the number of plays did not, that couldn’t have happened.

The second test case is a nice example of a Super Expert level.

In the third test case the number of plays decreased, which is impossible.

The fourth test case is probably an auto level with a single jump over the spike.

In the fifth test case the number of clears decreased, which is also impossible.

Nobody wanted to play the sixth test case; Polycarp’s mom attempted it to make him feel better, however, she couldn’t clear it.

思路:p(i+1) >= p(i), c(i+1) >= c(i)并且p(i) >= c(i),p(i) - p (i-1) >= c(i) - c(i-1)若不满足,则返回false.

#include <iostream>
using namespace std;int main()
{int T;scanf("%d", &T);while(T--){int n;bool flag = true;scanf("%d", &n);int p[1010]  = {0}, c[1010] = {0}, maxp = 0, maxc = 0;for(int i = 1; i <= n; i++){int zp = 0, zc = 0;scanf("%d%d", &p[i], &c[i]);if(p[i] > maxp){zp = p[i] - maxp;maxp = p[i];}else if(p[i] < maxp)flag = false;if(c[i] > maxc){zc = c[i] - maxc;maxc = c[i];}else if(c[i] < maxc)flag = false;if(p[i] >= c[i] && maxp >= maxc && zp >= zc)continue;elseflag = false;}if(flag)printf("YES\n");elseprintf("NO\n");}return 0;
}

这篇关于cf Educational Codeforces Round 85 (Rated for Div. 2)A. Level Statistics的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!



http://www.chinasem.cn/article/678887

相关文章

Codeforces Round #240 (Div. 2) E分治算法探究1

Codeforces Round #240 (Div. 2) E  http://codeforces.com/contest/415/problem/E 2^n个数,每次操作将其分成2^q份,对于每一份内部的数进行翻转(逆序),每次操作完后输出操作后新序列的逆序对数。 图一:  划分子问题。 图二: 分而治之,=>  合并 。 图三: 回溯:

cf 164 C 费用流

给你n个任务,k个机器,n个任务的起始时间,持续时间,完成任务的获利 每个机器可以完成任何一项任务,但是同一时刻只能完成一项任务,一旦某台机器在完成某项任务时,直到任务结束,这台机器都不能去做其他任务 最后问你当获利最大时,应该安排那些机器工作,即输出方案 具体建图方法: 新建源汇S T‘ 对任务按照起始时间s按升序排序 拆点: u 向 u'连一条边 容量为 1 费用为 -c,

Codeforces Round #261 (Div. 2)小记

A  XX注意最后输出满足条件,我也不知道为什么写的这么长。 #define X first#define Y secondvector<pair<int , int> > a ;int can(pair<int , int> c){return -1000 <= c.X && c.X <= 1000&& -1000 <= c.Y && c.Y <= 1000 ;}int m

Codeforces Beta Round #47 C凸包 (最终写法)

题意慢慢看。 typedef long long LL ;int cmp(double x){if(fabs(x) < 1e-8) return 0 ;return x > 0 ? 1 : -1 ;}struct point{double x , y ;point(){}point(double _x , double _y):x(_x) , y(_y){}point op

Codeforces Round #113 (Div. 2) B 判断多边形是否在凸包内

题目点击打开链接 凸多边形A, 多边形B, 判断B是否严格在A内。  注意AB有重点 。  将A,B上的点合在一起求凸包,如果凸包上的点是B的某个点,则B肯定不在A内。 或者说B上的某点在凸包的边上则也说明B不严格在A里面。 这个处理有个巧妙的方法,只需在求凸包的时候, <=  改成< 也就是说凸包一条边上的所有点都重复点都记录在凸包里面了。 另外不能去重点。 int

CF 508C

点击打开链接 import java.util.Arrays;import java.util.Scanner;public class Main {public static void main(String [] args){new Solve().run() ;} }class Solve{int bit[] = new int[608] ;int l

Codeforces 482B 线段树

求是否存在这样的n个数; m次操作,每次操作就是三个数 l ,r,val          a[l] & a[l+1] &......&a[r] = val 就是区间l---r上的与的值为val 。 也就是意味着区间[L , R] 每个数要执行 | val 操作  最后判断  a[l] & a[l+1] &......&a[r] 是否= val import ja

CSS实现DIV三角形

本文内容收集来自网络 #triangle-up {width: 0;height: 0;border-left: 50px solid transparent;border-right: 50px solid transparent;border-bottom: 100px solid red;} #triangle-down {width: 0;height: 0;bor

创建一个大的DIV,里面的包含两个DIV是可以自由移动

创建一个大的DIV,里面的包含两个DIV是可以自由移动 <body>         <div style="position: relative; background:#DDF8CF;line-height: 50px"> <div style="text-align: center; width: 100%;padding-top: 0px;"><h3>定&nbsp;位&nbsp;

Codeforces Round 971 (Div. 4) (A~G1)

A、B题太简单,不做解释 C 对于 x y 两个方向,每一个方向至少需要 x / k 向上取整的步数,取最大值。 由于 x 方向先移动,假如 x 方向需要的步数多于 y 方向的步数,那么最后 y 方向的那一步就不需要了,答案减 1 代码 #include <iostream>#include <algorithm>#include <vector>#include <string>