2021-01-30_nezzar has nn balls, numbered with integers 1, 2,

11：00~11：30 背单词

今天睡的太死了，闹钟都叫不醒我，醒来一看手机发现已经十点了，唉

14：00~16：00 看csdn

19：00~21：30 测试

本来还想把下午的时间用来刷题，没想到测试提前了，刷了两个题的我有点菜菜的。

You are given two integers n and k.

You should create an array of n positive integers a1,a2,…,an such that the sum (a1+a2+⋯+an) is divisible by k and maximum element in a is minimum possible.

What is the minimum possible maximum element in a?

Input
The first line contains a single integer t (1≤t≤1000) — the number of test cases.

The first and only line of each test case contains two integers n and k (1≤n≤109; 1≤k≤109).

Output
For each test case, print one integer — the minimum possible maximum element in array a such that the sum (a1+⋯+an) is divisible by k.

Example
Input
4
1 5
4 3
8 8
8 17
Output
5
2
1
3
Note
In the first test case n=1, so the array consists of one element a1 and if we make a1=5 it will be divisible by k=5 and the minimum possible.

In the second test case, we can create array a=[1,2,1,2]. The sum is divisible by k=3 and the maximum is equal to 2.

In the third test case, we can create array a=[1,1,1,1,1,1,1,1]. The sum is divisible by k=8 and the maximum is equal to 1.

这道题是个签到题，直接上代码吧。

#include<stdio.h>

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        long long n,k,a,b=0;
        scanf("%lld%lld",&n,&k);
        if(n>k)
        {
            if(n%k!=0) a=n/k+1;
            else a=n/k;
            k*=a;
        }
        a=k/n;
        if(k>a*n) b++;
        printf("%lld\n",a+b);
    }
}

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Nezzar has n balls, numbered with integers 1,2,…,n. Numbers a1,a2,…,an are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that ai≤ai+1 for all 1≤i<n.

Nezzar wants to color the balls using the minimum number of colors, such that the following holds.

For any color, numbers on balls will form a strictly increasing sequence if he keeps balls with this chosen color and discards all other balls.
Note that a sequence with the length at most 1 is considered as a strictly increasing sequence.

Please help Nezzar determine the minimum number of colors.

Input
The first line contains a single integer t (1≤t≤100) — the number of testcases.

The first line of each test case contains a single integer n (1≤n≤100).

The second line of each test case contains n integers a1,a2,…,an (1≤ai≤n). It is guaranteed that a1≤a2≤…≤an.

Output
For each test case, output the minimum number of colors Nezzar can use.

Example
Input
5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1
Output
3
2
4
1
1
这道题其实我题目没怎么看明白，但是样例很清楚，也是个难度不高的题

#include<stdio.h>

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,a[1010],max=0,x=1,i=0;
        scanf("%d",&n);
        for(i=0; i<n; i++)
            scanf("%d",&a[i]);
        for(i=0; i<n; i++)
        {
            //printf("%d\n",a[i]);
            if(i!=0)
            {
                //printf("x=%d\n",x);
                if(a[i]==a[i-1])
                    x++;
                else
                {
                    max=(max>x)?max:x;
                    x=1;
                }
            }
        }
        max=(max>x)?max:x;
        printf("%d\n",max);
    }
}

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21：30~23：30 刷题

问题 B: 二叉树，知中序后序，求先序（递归）
描述
已知二叉树的中序、后序，建树求先序。

格式
输入格式
中序序列
后序序列

输出格式
先序序列

样例
样例输入 Copy
d g b a e c f
g d b e f c a
样例输出 Copy
a b d g c e f

今天一直卡在这道题上，唉。后面发现是输入的问题，卡在这里真不值
代码：

#include<stdio.h>
#include<stdlib.h>
#define LEN sizeof(struct TREE)

struct TREE
{
    char data;
    struct TREE *l,*r;
};
typedef struct TREE A;

A * fun1(char ch1[],char ch2[],int n)
{
    A *p;
    if(n==0)
    {
        p=NULL;
    }
    else
    {
        int i=0,n1,n2;
        char x=ch2[n-1];
        A *q1,*q2;
        p=(A *)malloc(LEN);
        p->data=x;
        for(i=0; i<n; i++)
            if(ch1[i]==x)
                break;
        n1=i,n2=n-n1-1;
        //printf("n1=%d n2=%d\n",n1,n2);
        q1=fun1(ch1,ch2,n1);
        q2=fun1(ch1+i+1,ch2+n1,n2);
        p->l=q1,p->r=q2;
    }
    return p;
}
void fun2(A * p)
{
    if(p!=NULL)
    {
        printf("%c ",p->data);
        fun2(p->l);
        fun2(p->r);
        free(p);
    }
}
int main()
{
    char ch[1010];
    int n=0;
    while(scanf("%s",&ch[n++])!=EOF);
    //printf("%s",ch);
    A *root;
    root=fun1(ch,ch+n/2,n/2);
    fun2(root);
}

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总计学习7小时。
测试完人都不好了，今天早点睡，明天早点起，拜拜。