此题无法用JavaAC,不相信的可以去HD1029题试下! 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 hav…

方法一:    直接进行排序,输出第(n+1)/2位置上的数即可. (容易超时,关闭同步后勉强卡过) #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int n; ]; int main() { ios::sync_with_stdio(false); while(cin&g…

HDU 1029 题目大意:给定数字n(n <= 999999 且n为奇数 )以及n个数,找出至少出现(n+1)/2次的数 解题思路:n个数遍历过去,可以用一个map(也可以用数组)记录每个数出现的次数, 若次数一旦达到(n+1)/2,即输出a[i] 注意能出现(n+1)/2次数的最多只有一个 /* HDU 1029 *Ignatius and the Princess IV --- dp */ #include <cstdio> #include <cstring> #in…

HDU 1029 Ignatius and the Princess IV (思维题,排序?) 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 t…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1029 Ignatius and the Princess IV Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)Total Submission(s): 42359    Accepted Submission(s): 18467 Problem Description "OK, y…

Ignatius and the Princess IV Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)Total Submission(s): 27227    Accepted Submission(s): 11562 Problem Description "OK, you are not too bad, em... But you can never pass the…

Ignatius and the Princess IV Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32767K (Java/Other) Total Submission(s) : 7   Accepted Submission(s) : 3 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description "OK, you a…

Ignatius and the Princess IV Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32767 K (Java/Others)Total Submission(s): 51503    Accepted Submission(s): 23178 Problem Description "OK, you are not too bad, em... But you can never pass the…

题目链接:hdu 1028 Ignatius and the Princess III 题意:对于给定的n,问有多少种组成方式 思路:dp[i][j],i表示要求的数,j表示组成i的最大值,最后答案是dp[i][i].那么dp[i][j]=dp[i][j-1]+dp[i-j][i-j],dp[i][j-1]是累加1到j-1的结果,dp[i-j][i-j]表示的就是最大为j,然后i-j有多少种表达方式啦.因为i-j可能大于j,这与我们定义的j为最大值矛盾,所以要去掉大于j的那些值 /*******…

解题报告: 题目大意:就是要求输入的N个数里面出现的次数最多的数是哪一个,水题.暴力可过,定义一个一位数组,先用memset函数初始化,然后每次输入一个数就将下标对应的上标对应的那个数加一,最后将整个数组都扫一遍, 看哪一个最大就可以了. #include<cstdio> #include<cstring> ]; int main() { int N,d; while(scanf("%d",&N)!=EOF) { memset(times,,sizeof…