标签：line pos gic i++ this 判断 cond ssi das
In a magical forest, there exists N bamboos that don‘t quite get cut down the way you would expect.
Originally, the height of the ith bamboo is equal to hi. In one move, you can push down a bamboo and decrease its height by one, but this move magically causes all the other bamboos to increase in height by one.
If you can do as many moves as you like, is it possible to make all the bamboos have the same height?
The first line of input is T – the number of test cases.
The first line of each test case contains an integer N (1 ≤ N ≤ 105) - the number of bamboos.
The second line contains N space-separated integers hi (1 ≤ hi ≤ 105) - the original heights of the bamboos.
OutputFor each test case, output on a single line "yes” (without quotes), if you can make all the bamboos have the same height, and "no" otherwise.
Example2
3
2 4 2
2
1 2
yes
no
思路：判断是否全为奇或偶
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include<cstdio> #include<string> #include<cstring> #include <stdio.h> #include <string.h> using namespace std; const int N = 1000009 ; int a[100009] ; int main() { int n ; cin >> n ; while(n--) { int m ; scanf("%d" , &m); int sum = 0 ; for(int i = 0 ; i < m ; i++) { scanf("%d" , &a[i]); } int flag = 1 ; if(m == 1) printf("yes\n"); else { for(int i = 1 ; i < m ; i++) { if((a[i] - a[i - 1])%2 != 0) { flag = 0 ; break; } } if(flag) printf("yes\n"); else printf("no\n"); } } return 0 ; }
标签：line pos gic i++ this 判断 cond ssi das
原文地址：https://www.cnblogs.com/nonames/p/11290742.html
