do
begin
if (a[i]=x )then begin find:=true; exit; end;
end;
find:=false;
end;
begin
count:=0;
readln(n);
for i:=1 to n do
read(a[i]);
// bubbleSort(a);

for i:=1 to n do
for j:=i+1 to n do
begin k:= a[i]+a[j];
if find(a,k)=true then inc(count);
end;
writeln(count);
readln(k);
end.

include<iostream> include<cstring>

using namespace std;
int a[11000];
bool an[44100];
int main()
{
int i,j,k,n,s=0;
cin>>n;
memset(an,false,sizeof(an));
for(i=1;i<=n;i++)
cin>>a[i];
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
for(k=j+1;k<=n;k++)
{
if(a[j]+a[k]==a[i]&&an[i]==false)
{
s++;
an[i]=true;
}
}
}
}
cout<<s;
return 0;
}

include<iostream> include<cstring>

using namespace std;
int a[11000];
bool an[44100];
int main()
{
int i,j,k,n,s=0;
cin>>n;
memset(an,false,sizeof(an));
for(i=1;i<=n;i++)
cin>>a[i];
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
for(k=j+1;k<=n;k++)
{
if(a[j]+a[k]==a[i]&&an[i]==false)
{
s++;
an[i]=true;
}
}
}
}
cout<<s;
return 0;
}

include<iostream> include<cstring>

using namespace std;
int a[11000];
bool an[44100];
int main()
{
int i,j,k,n,s=0;
cin>>n;
memset(an,false,sizeof(an));
for(i=1;i<=n;i++)
cin>>a[i];
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
for(k=j+1;k<=n;k++)
{
if(a[j]+a[k]==a[i]&&an[i]==false)
{
s++;
an[i]=true;
}
}
}
}
cout<<s;
return 0;
}

include<iostream> include<cstring>

using namespace std;
int a[11000];
bool an[44100];
int main()
{
int i,j,k,n,s=0;
cin>>n;
memset(an,false,sizeof(an));
for(i=1;i<=n;i++)
cin>>a[i];
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
for(k=j+1;k<=n;k++)
{
if(a[j]+a[k]==a[i]&&an[i]==false)
{
s++;
an[i]=true;
}
}
}
}
cout<<s;
return 0;
}

#include<iostream>
#include<cstring>
using namespace std;
int a[11000];
bool an[44100];
int main()
{
int i,j,k,n,s=0;
cin>>n;
memset(an,false,sizeof(an));
for(i=1;i<=n;i++)
cin>>a[i];
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
for(k=j+1;k<=n;k++)
{
if(a[j]+a[k]==a[i]&&an[i]==false)
{
s++;
an[i]=true;
}
}
}
}
cout<<s;
return 0;
}

水题一发
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int a[11000];
bool an[44100];
int main()
{
int i,j,k,n,s=0;
scanf("%d",&n);
memset(an,false,sizeof(an));
for(i=1;i<=n;i++)scanf("%d",&a[i]);
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
for(k=j+1;k<=n;k++)
{
if(a[j]+a[k]==a[i]&&an[i]==false)
{
s++;
an[i]=true;
}
}
}
}
printf("%d\n",s);
return 0;
}

神题目 楼下说的就是这个题目的坑人之处 1+4=5与2+3=5是一组！！！
无语了

题目有坑，1+4=5与2+3=5是一组！！！
我也被坑了
#include<stdio.h>
#include<stdlib.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);
//system("pause");
return 0;
}

//难度几乎为0，但一定要注意：问的是个数而不是方案数！！一定要注意。本人去年竞赛，就因为这个点没搞清楚，才痛失一等奖！

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
using namespace std;

int n,a[110],flag[10010],ans=0;

int cmp(const void *a,const void *b)
{
return *(int )a>(int *)b?1:-1;
}

int main()
{
scanf("%d\n",&n);
for(int i=1;i<=n;i++)
{
scanf("%d ",&a[i]);
}
qsort(a+1,n,sizeof(a[1]),cmp);
memset(flag,0,sizeof(flag));
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
for(int k=j+1;k<=n;k++)
{
if(a[i]+a[j]==a[k]&&a[i]!=a[k]) flag[a[k]]=1;
}
}
}
for(int i=1;i<=n;i++)
{
if(flag[a[i]]==1) ans++;
}
printf("%d\n",ans);
return 0;
}

无难度水题
var n,i,j,c,x:longint;
a,s:array[1..10000]of integer;
begin
read(n);
for i:=1 to n do read(a[i]);
for i:=1 to n do
for j:=i+1 to n do begin
inc(c);s[c]:=a[i]+a[j];
end;
for i:=1 to c do
for j:=1 to n do
if s[i]=a[j] then begin a[j]:=0;inc(x);break;end;
write(x);
end.

#include<cstdio>
#include<iostream>
using namespace std;
int n,a[110];
bool flag[10010];
int main()
{
int i,j,k,ans=0;
scanf("%d",&n);
for(i=1;i<=n;i++) scanf("%d",&a[i]);
for(i=1;i<=n;i++) flag[a[i]]=1;
for(i=1;i<=n;i++)
for(j=i+1;j<=n;j++)
if(a[i]+a[j]<=10000 && flag[a[i]+a[j]])
flag[a[i]+a[j]]=0,ans++;
printf("%d\n",ans);
return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;
int a[102];
bool t[200001];
int i,ans,n,j;
int main()
{
memset(t,0,sizeof(t));
a[0]=0;ans=0;
cin>>n;
for(i=1;i<=n;i++) cin>>a[i];
for(i=1 ;i<=n-1;i++)
for(j=i+1;j<=n;j++)
t[a[i]+a[j]]=1;
for(i=1;i<=n;i++)
if (t[a[i]]) ans=ans+1;
cout<<ans<<endl;
return 0;
}

简单枚举而已 亲注意 最好用longint不然 容易超
program shaoziwei;
var
a:array[1..200] of longint;
b:array[1..200] of longint;
temp,num,i,m,q,sign:**longint**;
begin
for i:=1 to 100 do
b[i]:=0;
sign:=0;
read(num);
for i:=1 to num do
read(a[i]);
for i:=1 to num-1 do
for m:=i+1 to num do
begin
temp:=a[i]+a[m];
for q:=1 to num do
if (temp=a[q])and(b[q]=0) then
begin
sign:=sign+1;
b[q]:=1;
break;
end;
end;
write(sign);
end.

注意两种问法：有几对数和为另一个数；
有几个数为另两个数的和。
#include<iostream>
#include<string.h>
using namespace std;
bool has[10003];
int a[103];
int main(){
//freopen("in.txt", "r", stdin);
int size;
cin >> size;
int i;
memset(has, 0, sizeof(has));
for (i = 0; i < size; i++){
cin >> a[i];
has[a[i]] = true;
}
int ans = 0;
int j;
for (i = 0; i < size; i++){
for (j = i+1; j < size; j++){
if (a[i] + a[j] <= 10000){
if (has[a[i] + a[j]]){
ans++;
has[a[i] + a[j]] = false;
}
}
}
}
cout << ans << endl;
return 0;
}

//其中有多少个数，恰好等于集合中另外两个（不同的）数之和？
//注意
//5
//1 2 3 4 5
//其中1+4=5,2+3=5，但5只能算作一次哦！！！
#include<iostream>
#include<string.h>

using namespace std;

int a[101];
bool flag[101];

int main()
{
int n, i, j, k, tot;

cin >> n;
memset(a,0,sizeof(a));
memset(flag,false,sizeof(flag));

tot = 0;

for(i = 1; i <= n; i++)
cin >> a[i];

for(i = 1; i < n; i++)
for(j = i+1; j <= n; j++)
{
int temp = a[i] + a[j];
for(k = 1; k <= n; k++)
{
if((k != i) && (k != j) && (temp == a[k])) flag[k] = true;
}
}

for (i = 1; i <= n; i++)
if(flag[i]) tot++;
cout << tot << endl;

return 0;
}

晕死，没看好题目，原来和等于集合的数那个数只能算一次，不是算组合个数而是算数个个数，脑残了实在是。
诶- -下次做题看来要细心点

因为数据很小，枚举的话绝对不会爆的。要是数据大了的话，这道题可以变得比较复杂！
###block code
program ex;
var n,i,ans,j,sum,k:longint;
data:array[1..100] of longint;
ps:array[1..100] of boolean;
begin
fillchar(ps,sizeof(ps),false);
read(n); ans:=0;
for i:=1 to n do
read(data[i]);

for i:=1 to n-1 do
for j:=i+1 to n do
begin
sum:=data[i]+data[j];
for k:=1 to n do
if (sum=data[k]) and (ps[k]=false) then
begin
inc(ans);
ps[k]:=true;
break;
end;
end;

write(ans);
end.

代码
#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;

return 0;
}

O(max*n)

Pascal Code

var
n,t,s,i,j:longint;
num:array[1..10000] of boolean;
num2:array[1..10000] of longint;
flag:array[1..10000] of boolean;
begin
readln(n);
for i:=1 to n do
begin
read(t);
num2[i]:=t;
num[t]:=true;
end;
fillchar(flag,sizeof(flag),0);
for i:=1 to n do
for j:=1 to num2[i]-1 do
if (num[j]) and (num[num2[i]-j]) and (num2[i]-j<>j) and not (flag[i]) then
begin
s:=s+1;
flag[i]:=true;
end;
writeln(s);
end.

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int a[100+5],n,ans;
bool ok[100];
int main()
{
memset(ok,false,sizeof(ok));
cin>>n;
for(int i=1;i<=n;i++){cin>>a[i];}
sort(a+1,a+n+1);
for(int i=n;i>=1;i--)
{
for(int j=1;j<=i;j++)
{
for(int k=j+1;k<=i;k++)
{
if(a[i]==a[j]+a[k]&&a[j]!=a[k]&&ok[i]==false){ans++;ok[i]=true;}
}
}
}
cout<<ans<<endl;
return 0;
}