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  • codeforce 1478A A. Nezzar and Colorful Balls 额?模拟 C

    codeforce 1478A A. Nezzar and Colorful Balls 额?模拟 C

    https://codeforces.com/contest/1478/problem/A.

    A. Nezzar and Colorful Balls
    time limit per test
    1 second
    memory limit per test
    512 megabytes
    input
    standard input
    output
    standard output

    Nezzar has nn balls, numbered with integers 1,2,,n1,2,…,n. Numbers a1,a2,,ana1,a2,…,an are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that aiai+1ai≤ai+1 for all 1i<n1≤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 11 is considered as a strictly increasing sequence.

    Please help Nezzar determine the minimum number of colors.

    Input

    The first line contains a single integer tt (1t1001≤t≤100) — the number of testcases.

    The first line of each test case contains a single integer nn (1n1001≤n≤100).

    The second line of each test case contains nn integers a1,a2,,ana1,a2,…,an (1ain1≤ai≤n). It is guaranteed that a1a2ana1≤a2≤…≤an.

    Output

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

    Example
    input
    Copy
    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
    Copy
    3
    2
    4
    1
    1
    
    Note

    Let's match each color with some numbers. Then:

    In the first test case, one optimal color assignment is [1,2,3,3,2,1][1,2,3,3,2,1].

    In the second test case, one optimal color assignment is [1,2,1,2,1][1,2,1,2,1].

     分析

    题目要找到任意的数据都可以形成严格的上升序列

    那么当一个数字重复出现的时候,一定要分配到一个数组里

    就看看所有数字最多重复出现了多少次就可以了

    代码

     https://codeforces.com/contest/1478/submission/105679337

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <vector>
    #include <math.h>
    #include <string.h>
    #include <algorithm>
    #include <iostream>
    #include <string>
    #include <sstream>
    #include <iostream>
    #include <time.h>
    #include <queue>
    #include <list>
    #include <map>
    #include <set>
    #include <vector>
    #include <stack>
    #include <string.h>
    #include <bitset>
    #define sf scanf
    #define pf printf
    #define lf double
    #define p123 printf("123
    ");
    #define pn printf("
    ");
    #define pk printf(" ");
    #define p(n) printf("%d",n);
    #define pln(n) printf("%d
    ",n);
    #define s(n) scanf("%d",&n);
    #define ss(n) scanf("%s",n);
    #define ps(n) printf("%s",n);
    #define sld(n) scanf("%lld",&n);
    #define pld(n) printf("%lld",n);
    #define slf(n) scanf("%lf",&n);
    #define plf(n) printf("%lf",n);
    #define sc(n) scanf("%c",&n);
    #define pc(n) printf("%c",n);
    #define gc getchar();
    #define ll long long
    #define re(n,a) memset(n,a,sizeof(n));
    #define len(a) strlen(a)
    #define eps 1e-13
    #define zero(x) (((x) > 0? (x):(-x)) < eps)
    using namespace std;
    int a[2005];
    int b[2005];
    int main(){
        int t;
        s(t)
        while(t --){
            int n;
            s(n)
            for(int i = 0; i < n; i ++){
                s(a[i])
            }
            sort(a,a+n);
            int maxi = 0;
            int temp = 1;
            for(int i = 1; i < n; i ++){
                if(a[i] == a[i-1]){
                    temp ++;
                }else{
                    maxi = max(maxi,temp);
                    temp = 1;
                }
            }
            maxi = max(maxi,temp);
            p(maxi) pn
        }
        return 0;
    }
    作者:kidgzz
    本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利。
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  • 原文地址:https://www.cnblogs.com/Kidgzz/p/14347332.html
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