program num;
var
i,ans,n,j,k:longint;
a:array [1..100] of longint;
g:array [1..10000] of longint;
procedure kp(l,r:longint);
var
i,j,m,t:longint;
begin
i:=l; j:=r;
m:=a[(l+r) div 2];
repeat
while a[i]<m do inc(i);
while a[j]>m do dec(j);
if i<=j then begin
t:=a[i]; a[i]:=a[j]; a[j]:=t;
inc(i); dec(j);
end;
until i>j;
if l<j then kp(l,j);
if i<r then kp(i,r);
end;
begin
fillchar(g,sizeof(g),0);
readln(n); ans:=0;
for i:=1 to n do read(a[i]);
kp(1,n);
for i:=1 to n-2 do
for j:=i+1 to n-1 do
for k:=j+1 to n do
if (a[i]+a[j]=a[k])and(g[a[k]]=0) then begin inc(g[a[k]]); inc(ans); end;
writeln(ans);
end.

#include<bits/stdc++.h>
using namespace std;
const int MAXN=111; 
int N,a[MAXN],vis[101000];
int main()
{
    cin>>N;
    for(int i=1;i<=N;i++)
        cin>>a[i];
    sort(a+1,a+1+N);
    int ans=0;
    for(int i=1;i<=N;i++)
        vis[a[i]]=1;
    for(int i=1;i<N;i++)
        for(int j=i+1;j<=N;j++)
            if(vis[a[i]+a[j]]==1)
            {
                vis[a[i]+a[j]]=0;//vis要清零
                ans++;
            }
    cout<<ans<<endl;
    return 0; 
}

亲测AC代码（NB指the bigger number）

#include <bits/stdc++.h>
using namespace std;
int n, i, j, sum, N[110], NB[100010];
int main() {
    cin >> n;

    for(i = 0; i < n; i++) {
        cin >> N[i];
        NB[N[i]] = 1;
    }

    for(i = 0; i < n; i++)
        for(j = i + 1; j < n; j++)
            NB[N[i] + N[j]] = 0;

    for(i = 0; i < n; i++)
        if(NB[N[i]] == 0)
            sum++;

    cout << sum;
    return 0;
}

#include<stdio.h>
#include<map>
using namespace std;
const int maxn = 105;
int a[maxn];
int main()
{
int n;
map<int, int> mp;

while (scanf("%d", &n) != EOF)
{
int i, j;
for (i = 0; i < n; i++)
scanf("%d", &a[i]);
for (i = 0; i < n; i++)
{
mp[a[i]] = 1;
}
for (i = 0; i < n; i++)
{
for (j = i + 1; j < n; j++)
{
mp[a[i] + a[j]] = 0;
}
}
int c = 0;
for (i = 0; i < n; i++)
{
if (mp[a[i]] == 0)
c++;
}
printf("%d\n", c);
}
return 0;
}

#include<stdio.h>
#include<map>
using namespace std;
const int maxn = 105;
int a[maxn];
int main()
{
int n;
map<int, int> mp;

while (scanf("%d", &n) != EOF)
{
int i, j;
for (i = 0; i < n; i++)
scanf("%d", &a[i]);
for (i = 0; i < n; i++)
{
mp[a[i]] = 1;
}
for (i = 0; i < n; i++)
{
for (j = i + 1; j < n; j++)
{
mp[a[i] + a[j]] = 0;
}
}
int c = 0;
for (i = 0; i < n; i++)
{
if (mp[a[i]] == 0)
c++;
}
printf("%d\n", c);
}
return 0;
}

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

inline void input(vector<int> &p, int n){
int i;
for(i = 0; i < n; ++i){
cin>>p[i];
}
}
inline int count(vector<int> &p, int n){
int i, j, k;
int c = 0;
char is;
for(k = 2; k < n; ++k){

is = 0;

for(i = 0; i < k; ++i){
if(p[k] > p[i] + p[k - 1]){
continue;

}
for(j = k - 1; j > i; --j){
if(p[k] < p[i] + p[j]){
continue;

}else{
if(p[k] > p[i] + p[j]){
break;

}else{
is = 1;
break;
}
}
}
if(is){
++c;
break;
}
}

}
return c;
}

int main(){
int n;

cin>>n;

vector<int> s(n, 0);
input(s, n);

sort(s.begin(), s.end());

int c = count(s, n);

cout<<c<<endl;

return 0;

}

#include"stdio.h"
int a[101];
int b[101];
int main()
{
int n,i,j,k,sum=0,l;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}

for(i=1;i<=n;i++)
{
for(j=i;j<=n;j++)
{
if(a[i]>a[j])
{
k=a[i];
a[i]=a[j];
a[j]=k;
}
}
}

for(i=1;i<=n;i++)
{
b[i]=a[i];
}

for(i=1;i<=n;i++)
{
for(j=i+1;j<=n;j++)
{

k=a[i]+a[j];
for(l=j+1;l<=n;l++)
{
if(k==b[l])
{
sum++;
b[l]=0;
}
}
}
}

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

不要和我比简洁
var a:array[1..100]of longint;
f:array[1..20000]of boolean;
n,i,j,s:longint;
begin
read(n);
fillchar(f,sizeof(f),false);
for i:=1 to n do
begin
read(a[i]);
for j:=i-1 downto 1 do
f[a[i]+a[j]]:=true;
end;
for i:=1 to n do
if f[a[i]] then inc(s);
write(s);
end.

#include <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <bitset>
#include <cassert>
#include <map>
#include <string>
#include <sstream>
#include <ctime>
using namespace std;
int z[10010];
bool a[10010];
int main()
{
int n,s=0;
memset(a,0,sizeof(a));
scanf("%d",&n);
for(int i=0;i<=n-1;i++)
{
scanf("%d",&z[i]);
}
sort(z,z+n);
for(int i=1;i<=n-1;i++)
{
for(int j=0;j<i;j++)
{
for(int k=i;k<=n-1;k++)
{
if(z[k]==z[i]+z[j])
{
if(a[k]==0)
{
a[k]=1;
s++;
}
}
}
}
}
printf("%d",s);
return 0;
}

pascal 暴力解决（题目太坑，做错了好几遍

var n,i,j,k,t,x,m:integer;
a:array[1..100]of integer;b:array[1..10000]of boolean;
begin
    readln(n);m:=0;
    fillchar(b,sizeof(b),true);
    for i:= 1 to n do begin 
        read(a[i]);
        if a[i]>m then m:=a[i];
    end;
    for i:= 1 to n do 
        for j:= i+1 to n do begin
        x:=a[i]+a[j];
        if (i<>j)and(x<=m)and(b[x]=true) then 
            for k:= 1 to n do if x=a[k] then begin 
                t:=t+1;
                b[x]:=false;
            end;
    end;
    write(t);
end.

pascal AC

var a,g:array[0..100001] of longint;
n,i,j:longint;
s:qword;
begin
readln(n);
for i:=1 to n do
begin
read(a[i]);
inc(g[a[i]]);
end;
for i:=1 to n-1 do
for j:=i+1 to n do
if g[a[i]+a[j]]=1 then begin inc(s); inc(g[a[i]+a[j]]); end;
write(s);
end.

怎么说呢，本来用这个代码的时候是int Number[n]的，结果次次Runtime Error,只好看了看题，改了100和100000,因为10000也报错……

#include <iostream>

using namespace std;

int main()
{
    int n,i,j,total=0;
    cin>>n;
    int Number[100]={0};
    int NumberBiger[100000]={0};
    for(i=0;i<n;i++)
    {
        cin>>Number[i];
        NumberBiger[Number[i]]=1;
    }
    
    for(i=0;i<n;i++)
        for(j=i+1;j<n;j++)
            NumberBiger[Number[i]+Number[j]]=0;
    
    for (i=0;i<n;i++)
        if(NumberBiger[Number[i]]==0)
            total++;
    
    cout<<total;
    
    return 0;
}

这道题的思路主要是如何处理重复的情况，比如1+2=3，2+1=3。可以选择这样一种方法，对于某个数，使用三层循环遍历。最外面一层是为了验证每个数，里面两层分别验证加数和被加数。对于每个数，只要找到一种情况比如对于3找到1+2，那么就强制退出内两层循环并继续验证下一个数。

#include<iostream>

int quantity();
int* init(int n);
int compute(int temp[], int n);

int main()
{
    int n = quantity();
    int* temp = init(n);
    int result = compute(temp, n);
    std::cout << result << std::endl;

    return 0;
}

int quantity()
{
    using namespace std;
    int n = 0;
    cin >> n;

    return n;
}

int* init(int n)
{
    using namespace std;

    int* temp = new int[n];
    for(int i = 0;i < n;i++)
    {
        cin >> temp[i];
    }

    return temp;
}

int compute(int temp[] ,int n)
{
    using namespace std;
    int count = 0;

    for(int i = 0;i < n;i++)
        for(int j = 0;j < n;j++)
        {
            bool HaltJ = true;
            for(int k = 0;k < n;k++)
            {
                bool HaltK = true;
                if(temp[i] >= 3 && temp[j] != temp[k] && (temp[j] + temp[k]) == temp[i])
                {
                    count++;
                    HaltJ = false;
                    HaltK = false;
                }
                if(!HaltK)
                {
                    break;
                }
            }
            if(!HaltJ)
            {
                break;
            }
        }

    return count;
}

#include <stdio.h>
//goto的妙用
int main()
{
    int n;
    scanf("%d",&n);
    int nums[n];
    int times=0;
    for(int i=0;i<n;i++)
    {
        scanf("%d",&nums[i]);
    }
    /*
    one:需要找到两个数的和为它
    two:寻找数1
    three:寻找数2
    */
    for(int one=0;one<n;one++)
    {
        for(int two=0;two<n;two++)
        {
            if(one!=two)
            {
                for(int three=0;three<n;three++)
                {
                    if(three!=two&&three!=one&&nums[one]==(nums[two]+nums[three]))
                    {
                            times++;
                            //我知道用goto不推荐，但确实很方便
                            goto out;
                    }
                }
            }
        }
        out:continue;
    }
    printf("%d",times);
}

用了STL

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

int main()
{
    int n;
    cin >> n;
    std::unordered_set<int> in;
    for (int i = 0; i < n; ++i)
    {
        int tmp;
        cin >> tmp;
        in.insert(tmp);
    }
    std::unordered_set<int> out;
    for (const auto i : in)
    {
        for (const auto j : in)
        {
            if (（i != j） && （in.find(i + j) != end(in)）)
            {
                out.insert(i + j);
            }
        }
    }
    std::cout << out.size() << std::endl;
}

注意一下题目是问和的个数，也就是说（2+4=6 和 3+3=6 还有 1+5=6 都只算一种），所以就用一个 f[] 数组来记录这个数是否被算过和了。

#include <cstdio>
int a[200],f[30010],ans,n;

int main(){
    scanf("%d",&n);
    for (int i=1; i<=n; i++) f[(scanf("%d",&a[i]),a[i])]=1;
    for (int i=1; i<n; i++)
        for (int j=i+1; j<=n; j++)
            ans+=f[a[i]+a[j]],f[a[i]+a[j]]=0;
    printf("%d",ans);
}

来一份逻辑清晰的代码，不过下面有更加酷炫代码可以参考。

编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 512 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 508 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 512 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 508 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 = 512 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 512 KiB, score = 10

测试数据 #7: Accepted, time = 15 ms, mem = 508 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 508 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 512 KiB, score = 10

Accepted, time = 15 ms, mem = 512 KiB, score = 100

#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
int main()
{
    int ch[101];
    int sh[101];
    memset(sh,0,sizeof(sh));
    int n;
    scanf("%d",&n);
    for(int i=0; i<n; i++)
        scanf("%d",&ch[i]);
    std::sort(ch,ch+n);
    int count=0;
    for(int i=0; i<n-2; i++)
        for(int j=i+1; j<n-1; j++)
            for(int k=j+1; k<n; k++)
                if(ch[i]+ch[j] == ch[k]&&sh[k] == 0)
                {
                    sh[k] = 1;
                    count++;
                }
    printf("%d\n",count);
    return 0 ;
}

分站
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应用

搜索
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P2141 珠心算测验
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题目提供者soha
标签 NOIp普及组 2014 云端评测
难度 普及-
提交 讨论 题解

最新讨论

80分，不知道哪里出问题了
怎么还有被加数之说……？！
最后一个超时
pj第一题60+查不出错的的戳…
忘记去重了
评测结果：Accepted
得分：　　100
提交时间：2016-12-22 18:58 耗时：0ms
内存：11878kb
编译信息

编译成功
没有编译信息
分点信息（鼠标移到方块上有详细信息）

#1
AC
0ms/11878kB

#2
AC
0ms/11878kB

#3
AC
0ms/11878kB

#4
AC
0ms/11878kB

#5
AC
0ms/11878kB

#6
AC
0ms/11878kB

#7
AC
0ms/11878kB
#8
AC
0ms/11878kB

#9
AC
0ms/11878kB

#10
AC
0ms/11878kB
测试数据下载

因为该记录状态（AC或CE等）不能提供数据下载。

洛谷免费提供该记录第一个非AC的输入输出数据下载；部分题目因为版权等原因，不开放数据下载。24小时内，蓝名允许下载一次（一对输入及输出算一次），绿名2次，橙名3次，红名4次。灰名暂时不能下载数据。

源代码

#include <bits/stdc++.h>
using namespace std;
typedef long long int64;

int x[109];
int main(){
ios_base::sync_with_stdio(false);
int n , flag , beta = 0;
cin >> n;
for (int i=1;i<=n;++i) cin >> x[i];
sort(x+1,x+1+n);
for (int i=1;i<=n;++i){
flag=0;
for (int j=1;j<i;++j){
for (int k=1;k<j;++k){
if (x[j]+x[k]==x[i] && x[j]!=x[k]) {
flag=1;break;
}
}
}
beta+=flag;
}
cout << beta << endl;
return 0;
}

枚举
c++
#include<cstdio>
#include<algorithm>
int num[150];
int main()
{
int n,ans=0;scanf("%d",&n);
for (int i=1;i<=n;++i)
scanf("%d",num+i);
std::sort(num+1,num+n+1);
for (int k=3;k<=n;++k)
{
int f=1;
for (int i=k-2;f&&i;--i)
for (int j=k-1;f&&j>i;--j)
if (num[i]+num[j]==num[k])
{f=0;++ans;}
}
printf("%d\n",ans);
return 0;
}


#include <iostream>
#include <cmath>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <climits>  
#include <algorithm>
#include <set>

using namespace std;


int main()
{
  int n;
  cin>>n;

  int sum=0;

  int a[105];
  memset(a,0,sizeof(a));

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

  set<int> s;

  for(int i=1; i<=n; i++)
  {
    for(int j=1; j<=n&&j!=i; j++){s.insert(a[i]+a[j]);}
  }
  

  for(int i=1; i<=n; i++)
  {
    if(s.find(a[i])!=s.end()){sum++;}
  }
  
  cout<<sum;

  return 0;
}