水爆了。。
var n,i,j,k,sum:longint;
a,b:array[1..100] of longint;
begin
read(n);
fillchar(b,sizeof(b),0);
for i:=1 to n do read(a[i]);
for i:=1 to n do
for j:=i+1 to n do
for k:=1 to n do
if (a[i]<>a[j]) and (a[i]+a[j]=a[k]) then b[k]:=1;
for i:=1 to n do inc(sum,b[i]);
writeln(sum);
end.

大水题。。。。

var
a:array[1..100] of longint;
b:array[1..10000] of boolean;
n,i,j,k,t:longint;
begin
readln(n);
for i:=1 to n do
read(a[i]);
for i:=1 to n do
for j:=(i+1) to n do
for k:=1 to n do

if a[i]+a[j]=a[k] then b[a[k]]:=true;
for i:=1 to n do
if b[a[i]]=true then inc(t);
writeln(t);
end.

#include<iostream>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
bool cmp(int a,int b)
{
if(a>b)return 1;
return 0;
}
int a[10000];
int main()
{
int n,ans=0;
cin>>n;
for(int i=0;i<n;i++)
cin>>a[i];
sort(a,a+n,cmp);
for(int i=0;i<n-1;i++)
{
int y=i,z=n;
while(a[y+1]!=a[z-1]&&a[y+1]!=0&&a[z-1]!=0)
{
if(a[i]==a[y+1]+a[z-1]){
ans++;

break;
}
else if(a[i]>a[y+1]+a[z-1])z--;
else if(a[i]<a[y+1]+a[z-1])y++;

}
}
cout<<ans<<endl;
system("pause");
return 0;
}

var n,i,j,k,z:longint;
c:array[1..100] of longint;
d:array[1..100] of longint;
begin
readln(n);
for i:=1 to n do read(c[i]);
fillchar(d,sizeof(d),0);
for i:=1 to n do
for j:=i+1 to n do
for k:=1 to n do
if (c[i]+c[j]=c[k]) then inc(d[k]);
for i:=1 to n do
if d[i]>0 then inc(z);
writeln(z);
end.

var a,b,c,d,e:longint;
k,j:array[1..100]of longint;
begin
read(a);
for b:=1 to a do
read(k[b]);
for b:=1 to a do
for c:=1 to a do
for d:=c+1 to a do
if k[c]+k[d]=k[b] then begin j[b]:=1; break; break; end;
for b:=1 to a do
if j[b]=1 then inc(e);
writeln(e);
end.

#include<bits/stdc++.h>
int a[110];
int n;
int num=0;   
int v[110];  
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
       scanf("%d",&a[i]);
    for(int i=1;i<n;i++)  
       for(int j=i+1;j<=n;j++) 
          for(int k=1;k<=n;k++)   
             if(k!=i&&k!=j&&a[i]+a[j]==a[k]&&v[k]==0)  
                {
                   num++;
                   v[k]=1;   
                }
    printf("%d",num);
    return 0;
}

#include <stdio.h>

int main()
{
    int n,i,k,m=0,flag=0;
    int a[10000],b[10000];
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(i=0;i<n-1;i++)
    {
        for(k=i+1;k<n;k++)
        {
            b[m]=a[i]+a[k];
            m++;
        }
    }
    for(i=0;i<n;i++)
    {
        for(k=0;k<m;k++)
        {
            if(a[i]==b[k])
            {
                flag++;
                break;
            }
        }
    }
    printf("%d",flag);
    return 0;
}

用哈希表保存两两相加的值，然后在循环一波找到就ans++;

#include<iostream>
#include<string>
#include<map>
using namespace std;
int n;
int arr[101];
int main()
{
    int ans = 0;
    //哈希表
    map<int, int>hashmap;
    cin.tie(0);
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> arr[i];
    }
    for (int i = 0; i < n-1; i++) {
        for (int j = i + 1; j < n; j++) {
            hashmap[arr[i] + arr[j]] = 1;  // 记录相加的值
        }
    }
    for (int i = 0; i < n; i++) {
        if (hashmap[arr[i]] == 1) {  //如果数组中的值能在哈希表里面找到就ans++
            ans++;
        }
    }
    cout << ans << endl;
    return 0;
}

看看下一张

#include <cstdio> // 这题可以用桶排序的思想
#include <iostream>
using namespace std;
int t[20005], g[20005]; // t是桶，t[i]表示的值为i的数在集合中两两相加的次数；g[i]表示值为i的数是否在集合中，1为在，0为不在
int main(){
int n;
int sum = 0, a[101];
scanf("%d",&n);
for(int i = 1; i <= n; i++){
scanf("%d",&a[i]);
g[a[i]] = 1; //在集合中赋值为1
}
for(int i = 1; i < n; i++){ //枚举
for(int j = i+1; j <= n; j++){
t[a[i]+a[j]]++;
}
}
for(int i = 1; i <= 20002; i++){
if(t[i] > 0 && g[i]) //判断是否满足条件
sum++;
}
printf("%d",sum);
return 0;

}

import java.util.Scanner;

public class Main {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int a = in.nextInt();
int sum[] = new int[a];
boolean num[] = new boolean[a];
int x = 0, y = 0, d = 0;
for (int i = 0; i < sum.length; i++)
sum[i] = in.nextInt();
for (int i = 0; i < sum.length; i++) {
for (int j = i; j < sum.length; j++) {
for (int z = 0; z < sum.length; z++) {
if (sum[i] + sum[j] == sum[z] && i != j && i != z && j != z&&(!num[z])) {
num[z]=true;
y++;
}
}
}
}
System.out.println(y);
}
}

#include<iostream>
#include<cstdio>
using namespace std;
int t[200005],g[200005];//t是桶，t[i]表示值为i的数在集合中两两相加出现了几次，g[i]表示值为i的数是否在集合中，1为在，0为不在
int n,a[105],ans;
int main(){
cin>>n;
for (int i=1;i<=n;i++){
cin>>a[i];//读入
g[a[i]]=1;//在集合中赋值为1
}
for (int i=1;i<n;i++){//枚举
for (int j=i+1;j<=n;j++){
t[a[i]+a[j]]++;//被加出来了
}
}
for (int i=1;i<=200002;i++){
if (t[i]>0&&g[i]) ans++;//判断是否满足，满足ans++
}
cout<<ans;
return 0;
}

有没有python做的

简单的模拟 加一次排序可以减少重复 代码如下

#include <cstdio>
#include <algorithm>
using namespace std;
int n, ans;
int a[101];
bool v[101];
int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++ i)
        scanf("%d", &a[i]);
    sort( a + 1, a + 1 + n);
    for(int i = 1; i <= n - 2; ++ i)
    for(int j = i + 1; j <= n - 1; ++ j)
    for(int k = j + 1; k <= n; ++ k)
    if(a[i] + a[j] == a[k] && v[k] == 0){
        ans ++;
        v[k] = 1;
    }
    printf("%d", ans);
    return 0;
}

#include<stdio.h>
int main()
{
int n,i,j,k,flag;
scanf ("%d",&n);
int a[n],count=0;
for (i=0;i<n;i++)
{
scanf ("%d",&a[i]);
}
for (i=0;i<n;i++)
{
flag=0;
for (j=0;j<n;j++)
{
if (j!=i)
for (k=0;k<n;k++)
{
if (k!=j&&k!=i)
{
if (a[i]==a[k]+a[j])
{
count++;
flag=1;
}
}
if (flag==1)
break;
}
if (flag==1)
break;
}
}
printf ("%d",count);
}

#include<iostream>
#include<cstdio>
using namespace std;
int t[200005],g[200005];//t是桶，t[i]表示值为i的数在集合中两两相加出现了几次，g[i]表示值为i的数是否在集合中，1为在，0为不在
int n,a[105],ans;
int main(){
cin>>n;
for (int i=1;i<=n;i++){
cin>>a[i];//读入
g[a[i]]=1;//在集合中赋值为1
}
for (int i=1;i<n;i++){//枚举
for (int j=i+1;j<=n;j++){
t[a[i]+a[j]]++;//被加出来了
}
}
for (int i=1;i<=200002;i++){
if (t[i]>0&&g[i]) ans++;//判断是否满足，满足ans++
}
cout<<ans<<endl;
return 0;
}

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
int a[105];
int b[105];
int m;
void input(int n)
{
int i;
for(i=1;i<=n;i++)
{
cin>>a[i];
}

}
void sort(int n){
int i;
int j;
int tmp;
for(i=1;i<=n-1;i++)
for(j=1;j<=n-i;j++)
{
if(a[j]>a[j+1])
{
tmp=a[j];
a[j]=a[j+1];
a[j+1]=tmp;

}
}
}
int look(int aim,int left,int right){
int mid=(left+right)/2;
while(left<right)
{

if(a[mid]>aim)
{
right=mid;
mid=(right+left)/2;
continue;
}

if(a[mid]<aim)
{

left=mid+1;
mid=(right+left)/2;
continue;

}
// cout<<"mid="<<mid<<endl;
if(a[mid]==aim )
return 1;
}
return 0;
}
int main()
{
int i;
int j;
int n;
int aim;
int num=0;
cin>>n;
///cout<<"n="<<n<<endl;
input(n);
sort(n);///输入以及排序OK；0
1
memset(b,1,sizeof(b));
///cout<<"数组a=";
///for(i=1;i<=n;i++)
///{
/// cout<<a[i]<<',';
/// }
///cout<<endl;
for(i=1;i<=n;i++)
for(j=i+1;j<=n;j++)
{
if(b[j]==0)
continue;
aim=a[j]-a[i];
///cout<<"aim="<<aim<<endl;
if((a[i]==aim) ||(aim<a[i]))
continue;
else
{
if(look(aim,i,j))
{
/// cout<<"true"<<endl;
num++;
/// cout<<"aim="<<aim<<endl;
/// cout<<"num="<<num<<endl;
b[j]=0;
}
}

}

cout<<num<<endl;
return 0;
PS:想得有点麻烦了。。。

#include<iostream>
#include<cstdio>
using namespace std;
bool c[100000]={false},ct[100000]={false};
int in[1000]={0};
int main()
{
int n,ans=0;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>in[i];
c[in[i]]=true;
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(j==i)
{
continue;
}
if(c[in[i]+in[j]]==true)
{
ct[in[i]+in[j]]=true;
}
}
}
for(int i=0;i<100000;i++)
{
if(ct[i]==true)
{
ans++;
}
}
cout<<ans;
return 0;
}

//居然错了好几次。。。自己真的是智商下线了lol
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

int n, num[105], revans[100005], cnt;

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> num[i];
    for (int i = 1; i <= n; i++) {
        for (int j = i + 1; j <= n; j++) {
            revans[num[i] + num[j]] += 1;
        }
    }
    for (int i = 1; i <= n; i++) {
        cnt += (revans[num[i]] > 0);
    }
    cout << cnt << endl;
    return 0;
}