这个题目想到的最少有5种解法，有些可行有些不可行
1.使用**桶排序**的思想，开一个和数字大小一致的数组；但是数字的范围是10^9，因此这里是**不可行**的。
2.使用**hash**，将数字进行hash，遇到冲突决定是将频率增加还是在链表后面挂一个新值。
3.使用**sort**排序，然后从前往后扫描一遍并记录重复次数。或者使用二分查找找到上界和下界进行相减。
4.使用**map**，key对应着数字，value表示出现频率，最后用迭代器遍历一遍即可。
5.书写**二叉排序树**，节点记录值得同时记录下出现的频率；最后进行一遍中序遍历

标准离散化算法

#include<iostream>
#include<algorithm>
using namespace std;
//呵呵
//大神万岁

//结构定义区
struct lsv{
    long long value;
    int time;
    lsv(){
        value=0;time=0;
    }
};
//全局变量区
long long num[200001],n;
lsv ls[10001];
//函数声明区

//主函数开始！
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>num[i];
    }
    sort(num+1,num+n+1);
    int under=1;
    num[0]=0;
    for(int i=1;i<=n;i++)
    {
        if(num[i]==num[i-1]) ls[under].time++;
        else if(num[i]!=num[i-1])
        {
            under++;
            ls[under].value=num[i];
            ls[under].time=1;
        }
    }
    for(int i=1;i<=under;i++)
    {
        if(ls[i].time!=0)
        {
            cout<<ls[i].value<<' '<<ls[i].time;
            cout<<endl;
        }
    }
    return 0;
}
//函数实现区

简单的map

#include<iostream>
#include<map>
using namespace std;
int main(){
    map<int,int> A;
    int n,a;
    cin>>n;
    for (int i=0;i!=n;++i) {cin>>a;A[a]++;}
    for (auto &i:A) cout<<i.first<<' '<<i.second<<endl;
}

简单的C++代码

#include<bits/stdc++.h> 
using namespace std;
const int maxn=2e5;
int n,t=1;
int a[maxn];
int main()
{
    ios::sync_with_stdio(false);
    cin>>n;
    for(int i=0;i<n;i++) cin>>a[i];
    sort(a,a+n);
    for(int i=0;i<n;i++)
    {
        if(a[i]==a[i+1]) t++;
        if(a[i]!=a[i+1]&&a[i-1]==a[i])
        {
            cout<<a[i-1]<<" "<<t<<endl;
            t=1;
        }
        if(a[i]!=a[i+1]&&a[i-1]!=a[i]) cout<<a[i]<<" "<<t<<endl;
    }
    
}

map的做法：

#include <iostream>
#include <map>
using namespace std;
map<long long, int> book;
int main() {
    ios::sync_with_stdio(0);
    int n;  cin >> n;
    for(int i = 1; i <= n; i++) {
        long long a;    
        cin >> a;
        book[a]++;
    }
    for(auto it = book.begin(); it != book.end(); it++)
        cout << it->first << ' ' << it->second << endl; 
    return 0;   
}

/*
平衡二叉树，自己找轮子
*/
#include <iostream> //[2007提高组-A]统计数字
#include <algorithm>
using namespace std;

typedef struct TNode{
    int Data;
    TNode *Left, *Right;
    int t;  //计数
    int Height;
} * AVLTree;

int GetHeight(AVLTree AVL)
{
    if(AVL)
        return AVL->Height;
    else
        return 0;
}

AVLTree SingleLeftRotation(AVLTree A)   //LL旋
{
    AVLTree B = A->Left;
    A->Left = B->Right;
    B->Right = A;
    A->Height = max(GetHeight(A->Left), GetHeight(A->Right)) + 1;
    B->Height = max(GetHeight(B->Left), GetHeight(A)) + 1;
    return B;
}
AVLTree SingleRightRotation(AVLTree A)  //RR旋
{
    AVLTree B = A->Right;
    A->Right = B->Left;
    B->Left = A;
    A->Height = max(GetHeight(A->Left), GetHeight(A->Right)) + 1;
    B->Height = max(GetHeight(B->Right), GetHeight(A)) + 1;
    return B;
}
AVLTree DoubleLeftRightRotation(AVLTree A)  //LR旋
{
    A->Left = SingleRightRotation(A->Left);
    return SingleLeftRotation(A);
}
AVLTree DoubleRightLeftRotation(AVLTree A)  //RR旋
{
    A->Right = SingleLeftRotation(A->Right);
    return SingleRightRotation(A);
}

AVLTree Insret(AVLTree T, int x)
{
    if(!T)
    {
        T = new TNode();
        T->Data = x;
        T->Height = 1;
        T->t = 1;
    }
    else if(x < T->Data)
    {
        T->Left = Insret(T->Left, x);
        if (GetHeight(T->Left) - GetHeight(T->Right) == 2)
            if(x < T->Left->Data)
                T = SingleLeftRotation(T);
            else
                T = DoubleLeftRightRotation(T);
    }
    else if(x > T->Data)
    {
        T->Right = Insret(T->Right, x);
        if(GetHeight(T->Left) - GetHeight(T->Right) == -2)
            if(x > T->Right->Data)
                T = SingleRightRotation(T);
            else
                T = DoubleRightLeftRotation(T);
    }
    else
        T->t++;
    T->Height = max(GetHeight(T->Left), GetHeight(T->Right)) + 1;
    return T;
}

void Print(AVLTree T)
{
    if(T)
    {
        Print(T->Left);
        cout << T->Data << " " << T->t << endl;
        Print(T->Right);
    }
}

int main()
{
    int n, x;
    AVLTree AVL = NULL;
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> x;
        AVL = Insret(AVL, x);
    }
    Print(AVL);

    return 0;
}

#include<cstdio>
#include<iostream>
using namespace std;
void quicksort(int,int);
int a[200005];
int main(){
int n,i,flag,total;
scanf("%d",&n);
for(i=0;i<n;i++) scanf("%d",&a[i]);
quicksort(0,n-1);
flag=a[0];
total=1;
for(i=1;i<n;i++) {
if(a[i]!=flag){
printf("%d %d\n",flag,total);
total=0;
}
flag=a[i];
total++;
}
printf("%d %d\n",flag,total);
return 0;
}
inline void quicksort(int left,int right){
int i=left,j=right,mid=a[(left+right)/2];
while(i<=j){
while(a[i]<mid) i++;
while(a[j]>mid) j--;
if(i<=j){
swap(a[i],a[j]);
i++;
j--;
}
}
if(left<j) quicksort(left,j);
if(i<right) quicksort(i,right);
}

#include <iostream>
using namespace std;
int main()
{
int n,i,j,b,a[10000][2],k,l=0;
cin>>n;
for(i=0;i<10000;i++) {a[i][0]=0;a[i][1]=0;}
a[0][0]=1500000001;
for(i=0;i<n;i++)
{
cin>>k;
for(j=0;j<=l;j++)
if(a[j][0]==k) {a[j][1]++;break;}else
if(a[j][0]>k)
{
l++;
for(b=l;b>j;b--) {a[b][0]=a[b-1][0];a[b][1]=a[b-1][1];}
a[j][0]=k;
a[j][1]=1;
break;
}
}
for(i=0;i<10000;i++) if(a[i][1]!=0) cout<<a[i][0]<<' '<<a[i][1]<<endl;
return 0;
}

map大法好

#include <iostream>
#include <map>

using namespace std;

int main()
{
    ios::sync_with_stdio(false);

    map<long long, int> mp;
    int n, m;

    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> m;
        mp[m]++;
    }

    for (auto i = mp.begin(); i != mp.end(); i++)
        cout << i->first << " " << i->second << endl;

    return 0;
}

#include<bits/stdc++.h>
int n;
int num[200010];
int main() {
    scanf("%d", &n);
    for(int i=0; i<n; i++) scanf("%lld", &num[i]);
    std::sort(num, num+n);
    int count=1;
    for(int i=0; i<n-1; i++)
        if(num[i]==num[i+1]) count++;
        else {
            printf("%d %d\n", num[i], count);
            count=1;
        }
    printf("%d %d", num[n-1], count);
    return 0;
}

一个垃圾的代码，容菜鸟向神犇学习。。。
```cpp
#include<iostream>
#include<cstring>
#define MAXN 200001
using namespace std;
long long a[MAXN];
void qsort(int l,int r)
{

int i,j,mid,p;
i=l;
j=r;
mid=a[(l+r)/2];

do
{
while (a[i]<mid) i++;
while (a[j]>mid) j--;

if (i<=j)
{

p=a[i];
a[i]=a[j];
a[j]=p;
i++;
j--;

}
}
while (i<=j);

if (l<j) qsort(l,j);

if (i<r) qsort(i,r);
}

int main()
{
freopen("count.in","r",stdin);
freopen("count.out","w",stdout);
int n,count=1;
memset(a,0,sizeof(a));
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i];
qsort(1,n);
for(int i=2;i<=n+1;i++)
{
if(a[i]!=a[i-1])
{
cout<<a[i-1]<<" "<<count<<endl;
count=1;
}
if(a[i]==a[i-1])
count++;
}
return 0;
fclose(stdin);
fclose(stdout);
}
```

#include<bits/stdc++.h>
#define max_n 200005
using namespace std;
int a[max_n],n,last,cnt;
int main()
{
    scanf("%d",&n);
    for(register int i=1;i<=n;i++) scanf("%d",&a[i]);
    sort(a+1,a+n+1);
    last=a[1];
    for(register int i=1;i<=n;i++)
    {
        if(a[i]!=last) 
        {
            printf("%d %d\n",last,cnt);
            last=a[i];
            cnt=0;
        }
        cnt++;
    }
    printf("%d %d",last,cnt);
    return 0;
}

#include<iostream>
#include<algorithm>
using namespace std;
long long a[200010];
struct node{
    long long num;
    int ci;
}q[200100];
int main()
{
    int n,i,now=0;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    sort(a+1,a+n+1);
    for(i=1;i<=n;i++)
    {
        if(a[i]!=a[i-1])
        {
            now++;
            q[now].num=a[i];
            q[now].ci=1;
        }
        else
        q[now].ci++;
    }
    for(i=1;i<=now;i++)
     cout<<q[i].num<<" "<<q[i].ci<<endl;
    return 0;
}

这。
当年的noip为什么这么水。
简单的两种解法：
1. STL的sort，当年是不是要自己敲不清楚。
2. STL的map。

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int MO=19870816;
int N;
int A[200005];
int main () {
    
    scanf("%d", &N);
    for(int i=1; i<=N; i++) scanf("%d", &A[i]);
    sort(A+1, A+N+1);   
    int a=MO, b=0;
    for(int i=1; i<=N+1; i++)   
        if (A[i]!=a) {
            if (i!=1) printf("%d %d\n", a, b);
            a=A[i], b=1;
        }   
        else b++;   
    return 0;
    
}

#include<bits/stdc++.h>
#define For(i,j,k) for(int i=j;i<=k;i++)
using namespace std;
int a[200001],n;
map<int,int>mp;
int main()
{
    scanf("%d",&n);
    For(i,1,n)  scanf("%d",&a[i]),mp[a[i]]++;
    sort(a+1,a+n+1);
    int len=unique(a+1,a+n+1)-a-1;
    For(i,1,len)    printf("%d %d\n",a[i],mp[a[i]]);
}

C++ STL 大法好。
map万岁

#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
#define reg register
int main()
{
map<long long,int>t;
reg int n,i;
scanf("%d",&n);
long long a[n];
for(i=0;i<n;i++)
{
scanf("%lld",&a[i]);
t[a[i]]++;
}
sort(a,a+n);
printf("%lld %d\n",a[0],t[a[0]]);
for(i=1;i<n;i++)
if(a[i]!=a[i-1]) printf("%lld %d\n",a[i],t[a[i]]);
return 0;
}


#这年头使java的人真不多啊。。。奉上本题第一份java题解。。
##不说了，好水啊，水的我想吐。。。
```java
import java.util.Arrays;
import java.util.Scanner;

public class Main {
public static void main(String[] args){
Scanner cin = new Scanner(System.in);

int n = cin.nextInt();

int[] a = new int[200001];

for(int i = 0 ; i < n ; i ++){
a[i] = cin.nextInt();
}

Arrays.sort(a,0,n);

int tmp = 0;

for(int i = 0 ; i < n ; i ++){
tmp ++;
if(a[i] != a[i+1]){
System.out.println(a[i] + " " + tmp);
tmp = 0;
}
}

cin.close();
}
}
```
评测结果
编译成功

测试数据 #0: Accepted, time = 93 ms, mem = 43012 KiB, score = 10
测试数据 #1: Accepted, time = 109 ms, mem = 43008 KiB, score = 10
测试数据 #2: Accepted, time = 140 ms, mem = 43008 KiB, score = 10
测试数据 #3: Accepted, time = 234 ms, mem = 43004 KiB, score = 10
测试数据 #4: Accepted, time = 328 ms, mem = 43008 KiB, score = 10
测试数据 #5: Accepted, time = 328 ms, mem = 43000 KiB, score = 10
测试数据 #6: Accepted, time = 453 ms, mem = 43008 KiB, score = 10
测试数据 #7: Accepted, time = 734 ms, mem = 43008 KiB, score = 10
测试数据 #8: Accepted, time = 750 ms, mem = 43016 KiB, score = 10
测试数据 #9: Accepted, time = 93 ms, mem = 43008 KiB, score = 10
Accepted, time = 3262 ms, mem = 43016 KiB, score = 100

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

int a[200005];

int main()
{
    int i,j,k,n;
    cin>>n;
    for(i=1;i<=n;i++)
    cin>>a[i];
    sort(a+1,a+n+1);
    int sum=0;  int last=a[1];
    for(i=1;i<=n+1;i++)
    {
        if(a[i]==last)  sum++;
        else{cout<<a[i-1]<<" "<<sum<<endl;sum=1;last=a[i];}
    }
    return 0;
}

显然，离散化带个log轻松过
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN=200005;
int n,q[MAXN],a[MAXN],Hash[MAXN],val[MAXN];
int query(int L,int R,int x)
{
int l=L,r=R;
while(l<r){
int mid=(l+r)>>1;
if(x<=a[mid])r=mid;
else l=mid+1;
}
return l;
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&q[i]);
a[i]=q[i];
}
sort(a+1,a+n+1);
for(int i=1;i<=n;i++){
int s=query(1,n,q[i]);
++Hash[s];val[s]=q[i];
}
for(int i=1;i<=200000;i++){
if(val[i]>0)printf("%d %d\n",val[i],Hash[i]);
}
return 0;
}

从逻辑上来讲是错误的，但是卡了一个错位差——tot的加减问题
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <map>
#include <vector>
#include <stack>
#include <queue>

using namespace std;

int n,a[300000];

int main(){
scanf("%d",&n);
for(int i=0;i<n;i++)scanf("%d",&a[i]);
sort(a,a+n);
for(int i=0;i<n;){
printf("%d ",a[i]);
int t=a[i],tot=0;
while(a[i]==t)i++,tot++;
printf("%d\n",tot);
}
}