列表

详情


NC14322. Is Sorted

描述

A sequence contains n integers, can you make a judgment of whether the sequence is strictly increasing.

输入描述

There are multiple test cases. The first line is an positive integer indicating the number of test cases.
For each test case:
Line 1. A positive integer n, standing for the number of elements in the sequence.
Line 2. This line contains n integers, a1, a2, ..., an(2<=n<=100, 0<=ai<=1000) separated by space.

输出描述

For each test case, output one line. If the sequence is strictly increasing, print "Yes", else print "No".

示例1

输入:

3
3
1 2 3
3
3 2 1
4
1 1 2 3

输出:

Yes
No
No

原站题解

上次编辑到这里,代码来自缓存 点击恢复默认模板

C++14(g++5.4) 解法, 执行用时: 3ms, 内存消耗: 356K, 提交时间: 2018-05-15 00:54:09

#include <bits/stdc++.h>

using namespace std;

int t;
int n;
int a[110];

int main(void)
{
    cin>>t;
    while (t--)
    {
        cin>>n;
        for (int i=1;i<=n;++i)
            cin>>a[i];
        cout<<((is_sorted(a+1,a+1+n) && unique(a+1,a+1+n)==a+1+n)?"Yes":"No")<<endl;
    }

    return 0;
}

C++ 解法, 执行用时: 4ms, 内存消耗: 352K, 提交时间: 2021-06-30 15:37:19

#include <stdio.h>
int main ()
{
	int n,m,i,k,a,b;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d%d",&m,&a);
		for(k=1,i=2;i<=m;i++)
		{
			scanf("%d",&b);
			if(b<=a) k=0;
			a=b;
		}
		if(k) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}

C(clang 3.9) 解法, 执行用时: 1ms, 内存消耗: 376K, 提交时间: 2020-07-31 20:34:07

# include <stdio.h>
int main(){
	int n,m,j;
	scanf("%d",&n);
	while(n--){
		scanf("%d\n",&m);
        int i=-1,k=1;
		while(m--){
			scanf("%d",&j);
			if(i>=j) k=0;
			i=j;
		}
		printf(k?"Yes\n":"No\n");
	}
	return 0;
}

Python3(3.5.2) 解法, 执行用时: 27ms, 内存消耗: 3424K, 提交时间: 2018-12-18 00:58:24

n=int(input())

for i in range(n):
    Null=input()
    a=list(map(int,input().split()))
    if all(x<y for x,y in zip(a,a[1:])):
        print("Yes")
    else:
        print("No")

上一题