####E - Ignatius and the Princess IV
“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?
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.
Output
For each test case, you have to output only one line which contains the special number you have found.
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
Sample Output
3
5
1
####分析：
因為有內(nèi)存限制，所以用幾個大數(shù)組儲存數(shù)據(jù)并且進行運算是行不通的（如：a[i]=k） ，因此改變策略：輸入的n個數(shù)中 每個數(shù)都看作是一個位置信息（如a[k]++），通過置零該數(shù)組解決循環(huán)問題
####代碼實現(xiàn)：

/* 錯誤代碼： #include<iostream> #include<cstdio> long a[1000000] = {0} ;using namespace std ;int main () {long n , i , d , ; //最好用配套的long while (scanf( "%d" , &n ) != EOF ){for ( i = 0 ; i < n ; i++ ){// 數(shù)組第一個循環(huán)后沒有被清空 導(dǎo)致第二個循環(huán)里的數(shù)組不全為0 cin >> d ;a[d]++ ;if ( a[d] >= (n+1)/2 ){cout << d << endl ;break ; // a[1000000] = {0} ; //在這里遍歷long不好使 }}}return 0 ;} */ //正確代碼： #include<iostream> #include<cstdio> #include<string.h> long a[1000000] = {0}; //在外面定義 在里面賦值 using namespace std ; int main() {long n,i,t,re; //配套long while(scanf("%ld",&n)!=EOF)//在int main 里定義大數(shù)組不好使 { memset(a,0,1000000); //memset 把數(shù)組a 前1000000個值賦值0 如果用在整型數(shù)組里 只能賦值0或-1. for(i=0;i<n;i++){scanf("%ld",&t);a[t]++;if(a[t]>=(n+1)/2){re = t ;} }printf("%ld\n",re);} } /* 教訓(xùn)： 1：由于long型數(shù)組太大 無法在int main 里初始化 這就導(dǎo)致了前一個循環(huán)完成后，在后一個循環(huán)里數(shù)組的數(shù)不全為0 ， 所以要用memset去初始化 2：如果不用re代替t 則可能出現(xiàn)的情況中 輸入的數(shù)字都相同 那么結(jié)果會提前輸出 所以要建立一個中間變量去儲存該值 */

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