求解释为什么RE？
#include<cstdio>
using namespace std;
int n,a[101],ans;
bool b[101];
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=1;j<=n;j++)
{
if(i!=j)
b[a[i]+a[j]]=true;
}
for(int i=1;i<=n;i++)
if(b[a[i]])
ans++;
printf("%d",ans);
return 0;
}

#include<iostream>
#include<algorithm>
using namespace std;
int num[101];
bool flag[101];
int main()
{
int a,result=0;
cin>>a;
for(int i=0;i<a;i++)
cin>>num[i];
sort(num+0,num+a);
for(int i=0;i<a-1;i++)
for(int j=i+1;j<a;j++)
{
int temp=num[i]+num[j];
for(int k=0;k<a;k++)
if(num[k]==temp&&!flag[k])
{
result++;
flag[k]=true;
}
}
cout<<result;
return 0;
}

#include <bits/stdc++.h>
using namespace std;

int a[1000],n;
bool b[100000];
int main()
{
cin>>n;
for (int i=0;i<n;i++)
cin>>a[i];
for (int i=0;i<n;i++)
for (int j=0;j<n;j++)
if (a[i]!=a[j])
b[a[i]+a[j]]=true;

int ans=0;
for (int i=0;i<n;i++)
if (b[a[i]])
ans++;
cout<<ans<<endl;
return 0;
}

简单

#include<iostream>
using namespace std;
int main(){
    int n,a[102],ans=0,i,j,k;
    cin>>n;
    for(i=0;i<n;++i){
        cin>>a[i];
    }
    for(i=0;i<n;++i){
        for(j=0;j<n;++j){
            for(k=0;k<n;++k){
                if(a[i]==a[j]+a[k]&&j!=k){
                    ++i;
                    j=k=0;
                    ++ans;
                }
            }
        }
    }
    cout<<ans;
    return 0;
}

C
#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n, i;
    scanf("%d",&n);
    int a[100];
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    int j, k, t, s = 0;
    for(i=0;i<n;i++)
    {
        for(j=0;j<n-1;j++)
        {
            t = 1;
            for(k=j+1;k<n;k++)
            {
                if(k!=i && j!=i && a[k]+a[j]==a[i])
                {
                    t=0;
                    break;
                }
            }
            if(t == 0)
            {
                s++;
                break;
            }
        }
    }
    printf("%d",s);

    return 0;
}

题目有坑，还是个大坑！！！5=2+3和5=1+4是算同一个的！！！！！！！！
var
n,i,j,k,ans:longint;
a,b:array[1..10000] of longint;
procedure qsort(l,r:longint);
var
i,j,mid,p:longint;
begin
i:=l;
j:=r;
mid:=a[(l+r) div 2];
repeat
while a[i]<mid do inc(i);
while a[j]>mid do dec(j);
if i<=j then begin
p:=a[i]; a[i]:=a[j]; a[j]:=p;
inc(i);
dec(j);
end;
until i>j;
if l<j then qsort(l,j);
if i<r then qsort(i,r);
end;
begin
readln(n);
for i:=1 to n do begin
read(a[i]);
end;
readln;
qsort(1,n);
for i:=3 to n do begin
for j:=1 to i-2 do begin
for k:=j+1 to i-1 do begin
if a[j]+a[k]=a[i] then begin
b[i]:=1;
end;
end;
end;
end;
for i:=1 to n do begin
if b[i]=1 then inc(ans);
end;
writeln(ans);
end.

评测结果
编译成功

测试数据 #0: Accepted, time = 15 ms, mem = 520 KiB, score = 10
测试数据 #1: Accepted, time = 15 ms, mem = 516 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 516 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 512 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 512 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 516 KiB, score = 10
测试数据 #6: Accepted, time = 15 ms, mem = 516 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 516 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 512 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 516 KiB, score = 10
Accepted, time = 45 ms, mem = 520 KiB, score = 100
代码
#include <stdio.h>
#include <string.h>

int main()
{
int n,mark=0;
scanf("%d",&n);
int group[n],max,t;
for(int p=0;p<n;p++)
scanf("%d",&group[p]);
for(max=group[0],t=1;t<n;t++)
{
if(group[t]>max)
max=group[t];
}

int box[max];
memset(box,0,sizeof(int)*max);
for(int k=0;k<n;k++)
box[group[k]-1]=1;

for(int f=0;f<n-1;f++)
{
for(int r=f+1;r<n;r++)
{
for(int g=0;g<n;g++)
{
if((box[group[g]-1]==1) && group[f]+group[r]==group[g])
{
mark++;
box[group[g]-1]=0;
break;
}
}
}
}

printf("%d",mark);
return 0;
}

评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 524 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 524 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 528 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 528 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 528 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 524 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 524 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 528 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 528 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 524 KiB, score = 10
Accepted, time = 0 ms, mem = 528 KiB, score = 100
代码
#include <bits/stdc++.h>
bool hash[20001];
int a[1001],n,ans = 0;
int main() {
  //freopen("count.in","r",stdin);
  //freopen("count.out","w",stdout);
  scanf("%d",&n);
  memset(hash,false,sizeof(hash));
  for (int i = 1;i <= n;i++)
    scanf("%d",&a[i]);
  for (int i = 1;i <= n;i++)
    for (int j = 1;j <= n;j++) {
      if (i == j) continue;
      hash[a[i]+a[j]] = true;
    }
  for (int i = 1;i <= n;i++)
    if (hash[a[i]]) ans++;
  printf("%d",ans);
  return 0;
}

#include <cstdio>
#include <cstring>
const int MAXN=20009;
int main(){
    int a[MAXN],ans=0,n;
    bool vis[MAXN];
    scanf("%d",&n);
    memset(vis,false,sizeof(vis));
    for (int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        for (int j=1;j<=i-1;j++)
            if (a[i]!=a[j]) vis[a[i]+a[j]]=true;
    }
    vis[1]=false;
    for (int i=1;i<=n;i++) if (vis[a[i]]) ans++;
    printf("%d",ans);
    return 0;
}

//发现符合条件后即可跳出，进入下一个数。
代码
#include<stdio.h>
int main()
{
    int n,i,j,k,a[101],re=0;
    scanf("%d",&n);
    for(i=0;i<n;i++)scanf("%d",&a[i]);
    for(i=0;i<n;i++)
    {
      for(j=0;j<n;j++)
          for(k=0;k<n;k++)
              if(a[i]==a[j]+a[k]&&j!=k){i++;j=0;k=0;re++;}
    }
    printf("%d",re);
    return 0;
}```

var
i,j,k,n,x:longint;
a:array[0..100] of longint;
b:array[0..100] of boolean;
procedure paixu;
var
i,j:integer;
begin

for i:=1 to n-1 do
for j:=n downto i+1 do
if a[j]<a[j-1] then
begin
a[0]:=a[j];
a[j]:=a[j-1];
a[j-1]:=a[0];
end;

end;

begin
readln(n);
for i:=1 to n do

read(a[i]);
paixu;
for i:=1 to n do
for j:=1 to n do
for x:=1 to n do
if (i<j) and (j<x) and (a[i]+a[j]=a[x])
then
begin
b[x]:=true;
end;
for i:=1 to 100 do
if b[i] then k:=k+1;
writeln(k);
readln;
readln;
end.

用map随便搞搞就好了
```
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;

int cnt=0,N;
int a[110];
map<int,bool> m;
int main(){
cin>>N;
for(int i=1;i<=N;++i){
cin>>a[i];
}
for(int i=1;i<=N;++i){
for(int j=1;j<=N;++j){
if(i!=j){
m[a[i]+a[j]]=true;
}
}
}
for(int i=1;i<=N;++i){
if(m[a[i]]){
cnt++;
}
}
cout<<cnt;
return 0;
}
```

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>

using namespace std;

int n,x,a[20008],ans;
bool b[30000];

int main()
{
    //ios::sync_with_stdio(0);
    cin>>n;
    for(int i=1;i<=n;i++)
      {
        cin>>x;
        a[x]++;
      }
    for(int i=1;i<=10000;i++)
     if(a[i])
      for(int j=i+1;j<=10000;j++)
      {
       if(a[j]&& b[i+j]==0 &&a[i+j])
       {
         //cout<<i+j<<endl;
         b[i+j]=1;
         ans++;
       }
       if(i+j>10000) break;
     }
    cout<<ans;
    return 0;
}

#include <cstdio>

int main(){
// freopen("in.txt","r",stdin);
int n,a[500],c2[101]={0},count=0;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++)
if(a[k]==a[i]+a[j]&&k!=i&&k!=j)
c2[k]++;
for(int i=1;i<=n;i++)
if(c2[i]>0)
count++;
printf("%d",count);
return 0;
}

测试数据 #0: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #1: Accepted, time = 15 ms, mem = 488 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 484 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 488 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 488 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 484 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 484 KiB, score = 10
Accepted, time = 30 ms, mem = 488 KiB, score = 100

C：

#include <stdio.h>

int main() {
int i, j, k, count=0;
int n, A[101], b[101];
scanf("%d", &n);
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[i] == A[j] + A[k]) {
b[i] = 1;
}
}
}
}
for(i=1; i<=n; i++) {
if(b[i] == 1) count++;
}
printf("%d", count);
return 0;
}

#include<iostream>
#include<cstdio>
using namespace std;

const int maxn = 100 + 5;
const int maxnum = 100000 + 10;
int n;
int ans;
int maxx;
int num[maxn];
bool vis[maxnum];
bool vis1[maxnum];

int main ()
{
    //freopen ("in.txt", "r", stdin);
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> num[i];
        vis1[num[i]] = true;
        maxx = max (maxx, num[i]);
    }
    for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if (i != j && num[i] + num[j] <= maxnum) vis[num[i]+num[j]] = true;
    for (int i = 0; i <= maxx; i++) {
        if (vis[i] && vis1[i]) ans++;
    }
    cout << ans;
    return 0;
}

#include<cstdio>
int hash[10010],a[120],n,ans;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        int q=a[i];
        hash[q]=1;
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<i;j++){
            int q=a[i]+a[j];
            if(hash[q]==1){
                ans++;
                hash[q]=0;
            }
        }
    printf("%d\n",ans);
    return 0;
}

为什么会 runtimeerror？？

我觉得这道题应该注意的是：“其中有多少个数，恰好等于集合中另外两个（不同的）数之和”， 而不是：“集合中有多少对数的和在集合中”。不认真读题吃亏啊！

钙氧氮碘钡钡

测试数据 #0: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 560 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 560 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 560 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 560 KiB, score = 10
#include<iostream>
using namespace std;
int ans;
int a[101];

int n;
bool p[101];
int main()
{
int i,j,k;
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++)
if(a[i]>a[j])
swap(a[i],a[j]);
for(i=1;i<=n-2;i++)
for(j=i+1;j<=n-1;j++)
for(k=j+1;k<=n;k++)
if(a[i]+a[j]==a[k])
p[k]=1;
for(i=1;i<=n;i++)
if(p[i]==1)
ans++;
cout<<ans;
}