LightOJ - 1148 Mad Counting(坑)

最新推荐文章于 2023-10-30 20:23:39 发布

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最新推荐文章于 2023-10-30 20:23:39 发布

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原文链接：http://www.cnblogs.com/soTired/p/5330991.html

本文介绍了一种通过统计调查结果来快速估算一个城镇最小可能人口数量的方法。此方法避免了逐一计数的繁琐过程，利用了人们对足球赛事支持倾向的统计信息。通过对调查回复的数据进行特定处理，可以有效地计算出估计的最小人口数。

Mad Counting

Time Limit: 500MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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Description

Mob was hijacked by the mayor of the Town "TruthTown". Mayor wants Mob to count the total population of the town. Now the naive approach to this problem will be counting people one by one. But as we all know Mob is a bit lazy, so he is finding some other approach so that the time will be minimized. Suddenly he found a poll result of that town where N people were asked "How many people in this town other than yourself support the same team as you in the FIFA world CUP 2010?" Now Mob wants to know if he can find the minimum possible population of the town from this statistics. Note that no people were asked the question more than once.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with an integer N (1 ≤ N ≤ 50). The next line will contain N integers denoting the replies (0 to 10⁶) of the people.

Output

For each case, print the case number and the minimum possible population of the town.

Sample Input

1 1 2 2

Sample Output

Case 1: 5

Case 2: 1

Source

Problem Setter: Muhammad Rifayat Samee

Special Thanks: Jane Alam Jan

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什么鬼？相等的话跳过只加被询问的人 ; 然后对最后相等的数进行处理相加 ;

#include <cstdio>
#include <algorithm>
#define N 51
int a[N];
using namespace std;
int main()
{
    int t; int Q=1; 
    scanf("%d", &t);
    while(t--)
    {
        int n;
        scanf("%d", &n);
        for(int i=0; i<n; i++)
            scanf("%d", &a[i]);
        sort(a, a+n);
        int cnt=0, sp=0;
        for(int i=0; i<n; i++)
        {
            if(i==0 || a[i]==a[i-1]) cnt++; 
            else
            {
                int k=(cnt+a[i-1]) / (a[i-1]+1);   
                sp=sp+k*(a[i-1] + 1);
                cnt=0; 
            }
        }
        sp=sp+(cnt+a[n-1])/(a[n-1]+1) *(a[n-1]+1);
        printf("Case %d: %d\n", Q++, sp);
    } 
    return 0;
}

转载于:https://www.cnblogs.com/soTired/p/5330991.html