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来源：互联网发布：汽车之家淘宝编辑：程序博客网时间：2024/05/17 02:48

问题描述
给定n个正整数，找出它们中出现次数最多的数。如果这样的数有多个，请输出其中最小的一个。
输入格式
输入的第一行只有一个正整数n(1 ≤ n ≤ 1000)，表示数字的个数。
输入的第二行有n个整数s1, s2, …, sn (1 ≤ si ≤ 10000, 1 ≤ i ≤ n)。相邻的数用空格分隔。
输出格式
输出这n个次数中出现次数最多的数。如果这样的数有多个，输出其中最小的一个。
样例输入
6
10 1 10 20 30 20
样例输出
10

//#include<stdio.h>//#include<stdlib.h>//#include<algorithm> //#include<cmath>#include <map>#include <set>#include <cmath>#include <ctime>#include <Stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>using namespace std;int main(){    int a,b,c,d;    scanf("%d",&a);    int arr[a+1];    for(int i=0;i<a;i++)    {        scanf("%d",&arr[i]);    }    sort(arr,arr+a);    for(int i=0;i<a;i++)    cout<<arr[i]<<endl;    int temp=arr[0],ans=0,max=-1,count=1;    for(int i=0;i<a;i++)    {        count=0;        int j=i;        for(;arr[j]==temp;count++,j++)        cout<<"count"<<count<<" "<<j<<" "<<arr[j]<<" "<<temp<<endl;        if(count>max||(count==max&&arr[i]<ans))        {        cout<<ans<<" "<<count<<endl;            ans=arr[i];            max=count;                cout<<ans<<" "<<count<<endl;        }        temp=arr[i];        cout<<"temp"<<" "<<temp<<endl;    }    cout<<ans<<endl;    return 0;}

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别人的

#include <map>#include <set>#include <cmath>#include <ctime>#include <Stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout << "[" << x << "]"#define FIN freopen("input.txt", "r", stdin)#define FOUT freopen("output.txt", "w+", stdout)using namespace std;typedef long long LL;typedef pair<int, int> PII;typedef vector<LL> vec;typedef vector<vec> mat;const int MX = 1e3 + 5;int A[MX];int main() {    int n;    scanf("%d", &n);    for(int i = 1; i <= n; i++) {        scanf("%d", &A[i]);    }    sort(A + 1, A + 1 + n);    int Max = 0, ans, l, r;    for(l = 1; l <= n; l = r + 1) {        for(r = l; r < n && A[l] == A[r + 1]; r++);        int cnt = r - l + 1;        if(cnt > Max || (cnt == Max && A[l] < ans)) {            Max = cnt; ans = A[l];        }    }    printf("%d\n", ans);    return 0;}

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