題目鏈接：

http://acm.hdu.edu.cn/showproblem.php?pid=1029

題目類型：

給一個奇數個列的數組，其中一定存在某個數字，該數字的個數是大于一半的，問這個數字是幾

解題思路：

1、sort一遍，直接輸出下標為n/2的數。

2、將數組開為1000010，然后將數組置0，讀入一個數，下標為該數的數值+1，最后進行一次循環，判斷那個數值最大，輸出即可。（當年還不會快排所以用的這個方法，親測AC）。

題目：

Ignatius and the Princess IV

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32767 K (Java/Others)
Total Submission(s): 31704????Accepted Submission(s): 13646

Problem Description "OK, you are not too bad, em... But you can never pass the next test." feng5166 says.

"I will tell you an odd number N, and then N integers. There will be a special integer among them, you have to tell me which integer is the special one after I tell you all the integers." feng5166 says.

"But what is the characteristic of the special integer?" Ignatius asks.

"The integer will appear at least (N+1)/2 times. If you can't find the right integer, I will kill the Princess, and you will be my dinner, too. Hahahaha....." feng5166 says.

Can you find the special integer for Ignatius?

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Input The input contains several test cases. Each test case contains two lines. The first line consists of an odd integer N(1<=N<=999999) which indicate the number of the integers feng5166 will tell our hero. The second line contains the N integers. The input is terminated by the end of file.

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Output For each test case, you have to output only one line which contains the special number you have found.

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Sample Input 5 1 3 2 3 3 11 1 1 1 1 1 5 5 5 5 5 5 7 1 1 1 1 1 1 1

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Sample Output 3 5 1

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# include <stdio.h> # include <string.h> # define N 1000010 int n,a[N],b[N];int main () {int i,ret,max;while(scanf("%d",&n)!=EOF){memset(b,0,sizeof(b));max=0; ret=0;for(i=0;i<n;i++){scanf("%d",&a[i]);b[a[i]]++;if(b[a[i]] >max){max=b[a[i]];ret=a[i];}}printf("%d\n",ret); }return 0; }

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轉載于:https://www.cnblogs.com/love-sherry/p/6941505.html

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