HDOJ 1029 Ignatius and the Princess IV-优快云博客

本文介绍了一种使用 C++ 编程语言解决 HDU 1029 题目的方法，该题目要求找出给定数组中出现次数最多的元素。通过使用 map 数据结构来维护每个元素的出现次数，并最终找到频数最高的元素。

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题意：求n个数中个数最多的数

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

思路：map维护即可

注意点：无

以下为AC代码：

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
10633741	2014-04-27 17:23:16	Accepted	1029	78MS	232K	289 B	C++	luminous11

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cctype>
#include <climits>
#include <iomanip>
#include <cstdlib>
#include <algorithm>
//#include <unordered_map>
//#include <unordered_set>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define RDI(a) scanf ( "%d", &a )
#define RDII(a, b) scanf ( "%d%d", &a, &b )
#define RDIII(a, b, c) scanf ( "%d%d%d", &a, &b, &c )
#define RS(s) scanf ( "%s", s )
#define PI(a) printf ( "%d", a )
#define PIL(a) printf ( "%d\n", a )
#define PII(a,b) printf ( "%d %d", a, b )
#define PIIL(a,b) printf ( "%d %d\n", a, b )
#define PIII(a,b,c) printf ( "%d %d %d", a, b, c )
#define PIIIL(a,b,c) printf ( "%d %d %d\n", a, b, c )
#define PL() printf ( "\n" )
#define PSL(s) printf ( "%s\n", s )
#define rep(i,m,n) for ( int i = m; i <  n; i ++ )
#define REP(i,m,n) for ( int i = m; i <= n; i ++ )
#define dep(i,m,n) for ( int i = m; i >  n; i -- )
#define DEP(i,m,n) for ( int i = m; i >= n; i -- )
#define repi(i,m,n,k) for ( int i = m; i <  n; i += k )
#define REPI(i,m,n,k) for ( int i = m; i <= n; i += k )
#define depi(i,m,n,k) for ( int i = m; i >  n; i += k )
#define DEPI(i,m,n,k) for ( int i = m; i >= n; i -= k )
#define READ(f) freopen(f, "r", stdin)
#define WRITE(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);
template <class T>
inline bool RD ( T &ret )
{
    char c;
    int sgn;
    if ( c = getchar(), c ==EOF )return 0; //EOF
    while ( c != '-' && ( c < '0' || c > '9' ) ) c = getchar();
    sgn = ( c == '-' ) ? -1 : 1;
    ret = ( c == '-' ) ? 0 : ( c - '0' );
    while ( c = getchar() , c >= '0' && c <= '9' ) ret = ret * 10 + ( c - '0' );
    ret *= sgn;
    return 1;
}
inline void PD ( int x )
{
    if ( x > 9 ) PD ( x / 10 );
    putchar ( x % 10 + '0' );
}
const double eps = 1e-10;
const int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };
struct node{
    int x, y, cnt;
    node(){}
    node( int _x, int _y ) : x(_x), y(_y) {}
    node( int _x, int _y, int _cnt ) : x(_x), y(_y), cnt(_cnt) {}
};
vector<int> num;
int main()
{
    int n;
    while ( RDI ( n ) != EOF ){
        num.clear();
        int k;
        rep ( i, 0, n ){
            RD ( k );
            num.push_back ( k );
        }
        sort ( all( num ) );
        int cnt = 1;
        k = 1;
        int ans = num[0];
        rep ( i, 1, num.size() ){
            if ( num[i] == num[i-1] ){
                cnt ++;
            }
            if ( num[i] != num[i-1] ){
                if ( k < cnt ){
                    k = cnt;
                    ans = num[i-1];
                }
                cnt = 1;
            }
        }
        if ( k < cnt ){
            k = cnt;
            ans = num[num.size()-1];
        }
        PIL ( ans );
    }
    return 0;
}