BestCoder Round #50 (div.2) & HDOJ5364 Search Result(模拟)

原创于 2015-08-09 10:33:55 发布 · 1k 阅读

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本文探讨了一种解决特定问题的模拟算法，通过输入不同数量的整数和对应的ID，输出需要被惩罚的人的ID。当没有人需要被惩罚时，输出-1。详细解释了算法的实现过程和关键步骤。

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Input

There are multiply cases.
For each case,there is a single integer n(1<=n<=1000) in first line.
In second line,there are n integer a1,a2...an(0<=ai<10000)ai is the the ith man's ID.

Output

Output ID of the man who should be punished.
If nobody should be punished,output -1.

Sample Input

Sample Output

1
-1

简单的小模拟(竟然wa了。。读题啊！)

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 10005;
int n, a[maxn];
int main()
{
    while(scanf("%d", &n) != EOF) {
        memset(a, 0, sizeof(a));
        for(int i = 0; i < n; ++i) {
            int x;
            scanf("%d", &x);
            a[x]++;
        }
        int m = 0, flag, sum = 0;
        for(int i = 0; i < maxn; ++i) {
            if(a[i] > m) {
                m = a[i];
                flag = i;
            }
        }
        if(m == 1) {
            printf("-1\n");
            continue;
        }
        for(int i = 0; i < maxn; ++i)
            if(i != flag)
                sum += a[i];
        if(m > sum) printf("%d\n", flag);
        else printf("-1\n");
    }
    return 0;
}