LightOJ 1148 - Mad Counting【数学】

柔情只为你懂 2022-08-18 02:28 30阅读 0赞

1148 - Mad Counting

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   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Time Limit:&nbsp;<span style="color:rgb(180,95,4)">0.5 second(s)</span></td> 
   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Memory Limit:&nbsp;<span style="color:rgb(180,95,4)">32 MB</span></td> 
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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.

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   <td> <h1 style="color:green">Sample Input</h1> </td> 
   <td> <h1 style="color:green">Output for Sample Input</h1> </td> 
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   <td> <p>2</p> <p>4</p> <p>1 1 2 2</p> <p>1</p> <p>0</p> </td> 
   <td> <p>Case 1: 5</p> <p>Case 2: 1</p> </td> 
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解题思路

将说的数和说的次数同时存储，将每个数除以每个数说的次数向上取整相加的和，可以再输入时就将此数加一。

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    struct nn
    {
    	int shu1,wei;
    }num1[100001];
    struct node
    {
    	int shu;
    	int ci;
    }num[100001];
    int ans;
    int find(int x)
    {
    	int i;
    	for(i=0;i<ans;i++)
    	{
    		if(num[i].shu==x+1)
    		{
    			num[i].ci++;
    			return i;
    		}
    	}
    	if(i==ans)
    	{
    		num[i].shu=x+1;
    		num[i].ci++;
    		ans++; 
    		return i;
    	}
    } 
    int main()
    {
    	int t;
    	int wc=0;
    	scanf("%d",&t);
    	while(t--)
    	{
    		wc++;
    		ans=0;
    		memset(num,0,sizeof(num));
    		int n;
    		scanf("%d",&n);
    		for(int i=0;i<n;i++)
    		{
    			scanf("%d",&num1[i].shu1);
    			int cnm;
    			cnm=find(num1[i].shu1);
    			num1[i].wei=cnm;
    		}
    		int cc=0;
    		for(int i=0;i<ans;i++)
    		{
    			cc+=((num[i].shu+num[i].ci-1)/num[i].shu)*num[i].shu;
    		}
    		printf("Case %d: %d\n",wc,cc);
    	}
    	return 0;
    }