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  • Codeforces Round #351 Div2 C 简单枚举

    C. Bear and Colors
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Bear Limak has n colored balls, arranged in one long row. Balls are numbered 1 through n, from left to right. There are n possible colors, also numbered 1 through n. The i-th ball has color ti.

    For a fixed interval (set of consecutive elements) of balls we can define a dominant color. It's a color occurring the biggest number of times in the interval. In case of a tie between some colors, the one with the smallest number (index) is chosen as dominant.

    There are non-empty intervals in total. For each color, your task is to count the number of intervals in which this color is dominant.

    Input

    The first line of the input contains a single integer n (1 ≤ n ≤ 5000) — the number of balls.

    The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ n) where ti is the color of the i-th ball.

    Output

    Print n integers. The i-th of them should be equal to the number of intervals where i is a dominant color.

    Examples
    Input
    4
    1 2 1 2
    
    Output
    7 3 0 0 
    
    Input
    3
    1 1 1
    
    Output
    6 0 0 
    
    Note

    In the first sample, color 2 is dominant in three intervals:

    • An interval [2, 2] contains one ball. This ball's color is 2 so it's clearly a dominant color.
    • An interval [4, 4] contains one ball, with color 2 again.
    • An interval [2, 4] contains two balls of color 2 and one ball of color 1.

    There are 7 more intervals and color 1 is dominant in all of them.

    题意:

    每个球对应一种颜色,一共有n个球,所以有个区间,问每种颜色在多少个区间中是dominant?

    dominant的意思是在这个区间中这种颜色数最多,如果有相同个数的颜色,编号最小的是dominant。

    分析:

    枚举个区间,对于每个区间统计那种颜色是dominant即可。


    #include<bits/stdc++.h>
    using namespace std;
    const int N=5009;
    int ans[N],num[N],a[N];
    int main()
    {
        int n;scanf("%d",&n);
        for(int i=0;i<n;i++)scanf("%d",&a[i]);
        int tot=0;
        for(int i=0;i<n;i++){
            memset(num,0,sizeof(num));
            int maxn=0,maxp=i;
            for(int j=i;j<n;j++){
                tot++;
                num[a[j]]++;
                if(num[a[j]]>maxn||(a[j]<maxp&&num[a[j]]==maxn)){
                    maxn=num[a[j]];
                    maxp=a[j];
                    //ans[a[j]]++;
                }
                ans[maxp]++;
            }
        }
       // cout<<tot<<endl;
        for(int i=1;i<=n;i++)printf("%d ",ans[i]);
        return 0;
    }





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  • 原文地址:https://www.cnblogs.com/01world/p/5651219.html
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