python 暴力枚举

n = eval(input())
num = [eval(x) for x in input().split()]
lst = []
count = 0
for i in num:
    for j in num:
        if i==j:
            continue
        else:
            sum=i+j
            if sum in lst:
                continue
            else:
                lst.append(sum)
                if sum in num:
                    count+=1
print(count)
        

n = int(input())
num = list(input().split(' '))
sum = list()
ans = 0
k = list()
Temp = list()

for i in range(n):
    for j in range(n):
        if i != j:
            k.append(int(num[i])+int(num[j]))

for a in range(0,(n*n - n)):
    for b in range(0,n):
        if int(k[a]) == int(num[b]):
            if k[a] not in Temp:
                ans+=1
            Temp.append(k[a])
print(ans)

#include <stdio.h>

main() {

int N,n, x, l = 0;
scanf("%d %d", &n, &x);
for (int i = 1; i <= n; i++)
{
for (int j = 0,k=i; k >0; j++)
{

N = k % 10;
if (N == x)
l += 1;
k = k / 10;
}
}
printf("%d", l);
}

#include <iostream>         //[2014普及组-A]珠心算测验
#include <algorithm>
#include <map>
using namespace std;

int main()
{
    //map<int, int> m;
    int m[20001] = {0};
    int n, num, ans = 0;
    int N[100] = {0};
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> num;
        N[i] = num;
        m[num] = 1;
    }

    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)
            if(m[N[i] + N[j]] == 1)
            {
                m[N[i] + N[j]] = 0;
                ans++;
            }

    cout << ans << endl;

    return 0;
}

思路：数组计数，由于只记是否出现，用bitset压缩空间

#include <iostream>
#include <bitset>

using namespace std;

int main()
{
    bitset<20001> t;//其无参构造函数将全部元素置0，不需初始化
    int n, n_set[100], ans = 0;
    scanf("%d", &n);
    for (int i = 0; i < n; ++i)
        scanf("%d", &n_set[i]);
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < i; ++j)
            t[n_set[i] + n_set[j]] = 1;
    for (int i = 0; i < n; ++i)
        if (t[n_set[i]])++ans;
    printf("%d", ans);
}

#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;
int a[110]={};
int b[20002]={};

int main(){
int n;
cin>>n;
int *q;
int count=0;
for(int i=0;i<n;i++){
cin>>a[i];
b[a[i]]=1;
}
int x;
for(int j=0;j<n-1;j++){
for(int k=j+1;k<n;k++){
x=a[j]+a[k];
if(b[x]==1){
count++;
b[x]=0;
}
}
}
cout<<count<<endl;
return 0;
}

#include<stdio.h>
int main()
{
int n;
int i,b,c,a[100],count=0;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
for(b=0;b<n;b++)
for(c=0;c<n;c++)
if(b!=c && (a[i]==a[b]+a[c]))
{
count++;
i++;
b=0;c=0;
}
printf("%d",count);
return 0;
}

#include<iostream>
using namespace std;

int main()
{
int N;
cin >> N;
int *num = new int[N];
int a = 0;
int res = 0;
int temp = 0;
for (int i = 0; i < N; i++)
{
cin >> a;
num[i] = a;
}
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
for (int k = j+1; k < N; k++)
{
temp = res;
if ((num[i] == num[j] + num[k])&&(j!=k))//对num里的每个数进行遍历，为避免重复，（5 = 1+4，5=2+3），应跳出两重循环。
{
res++;
break;
}
}
if (res > temp)
{
break;
}
}
}
cout << res << endl;
return 0;
}

#include<iostream>
using namespace std;
int main(){
int n;
cin >> n;
int num;
num = (int)malloc(sizeof(int)*n);
for (int i = 0; i < n; i++){
cin >> num[i];
}
int lable = 0;
int current = 0;
while (current<=(n-1))
{
int limit = 0;
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
if (j != i&&j != current&&i != current&&limit==0){
if (num[i] + num[j] == num[current]){
lable++;
limit++;
cout << "hhh" << endl;
cout << " " << num[i] << " " << num[j] << " " << num[current]<<endl;
break;
}
}
}
}
current++;
}
cout << lable << endl;
return 0;
}

/*此算法虽然不高级，但是好理解，
就是建立一个数组b来存两数相加的值，然后遍历a[0]-a[n-1]
如有a[i]==数组b中的一个值，sum++;并且要break避免重复。
*/

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    int b[100002];
    int h=0;
    int i=0;
    int sum=0;
    int n;
    int j;
    cin>>n;
    int a[n];
    for(int k=0;k<n;k++)
    cin>>a[k];
    for(i=0;i<n;i++){
        for(j=i+1;j<n;j++){
            b[h]=a[i]+a[j];
            h++;
        }
    }
    for(j=0;j<n;j++){
        for(i=0;i<h;i++){
            if(a[j]==b[i]){
                sum++;
                break;
            }
        }
    }
    cout<<sum;
    return 0;
}

#include<iostream>
#include<algorithm>
using namespace std;
bool cmp(int a,int b)
{
return a<b;
}
int main()
{
int n;
cin>>n;
int *num = new int[n];
for(int i=0;i<n;i++)
{
cin>>num[i];
}
sort(num,num+n,cmp);
int cnt = 0;
for(int i=2;i<n;i++)
{
for(int j=0,k=i-1;j<k;)
{
if(num[j]+num[k]==num[i])
{
cnt++;
break;
}else if(num[j]+num[k]<num[i]) j++;
else k--;
}
}
cout<<cnt<<endl;
}

#include<iostream>
using namespace std;
int main(){
int number;
const int N=100;
long int test[N];
int op=0;
int i=0;
int u;
int count=0;
int iffind=0;
cin>>number;
for(;i<number;i++){
cin>>test[i];
}
for(i=0;i<number;i++){
iffind=0;
for(u=0;u<number-1;u++){
if(iffind==1) break;
for(op=u+1;op<number;op++){
if(op!=i&&u!=i&&test[i]==test[u]+test[op]){
count++;
iffind=1;
break;
}
}
}
}
cout<<count;
}

#include <iostream>
using namespace std;
int main()
{
int a[20002]={0};
bool visited[20002]={0};
int n;
cin>>n;
int x,sum=0;
int maxi = 0;
while(n--){
cin>>x;
a[x] = 1;
if(maxi<x)
maxi = x;
}
for(int i = 1;i < maxi; i++){
if(a[i]==1)
for(int j = i+1;j <= maxi; j++){
if(a[j]==1){
if(a[i+j]==1&&visited[i+j]!=1){
sum++;
visited[i+j] = 1;
}
}
}
}
cout<<sum;

}

c++代码
#include<bits/stdc++.h>
using namespace std;
int a[103],b[103];
int main(){
int n,s=0;
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++)
for(int q=1;q<=n;q++)
if(a[j]!=a[q] && a[j]+a[q]==a[i]) b[i]=1;
for(int i=1;i<=n;i++)
if(b[i]==1) s++;
cout<<s;
return 0;
}

#include<iostream>
#include<cstdio>
using namespace std;
int t[200005],g[200005];
int n,a[105],ans;
int main(){
cin>>n;
for (int i=1;i<=n;i++){
cin>>a[i];
g[a[i]]=1;
}
for (int i=1;i<n;i++){
for (int j=i+1;j<=n;j++){
t[a[i]+a[j]]++;
}
}
for (int i=1;i<=200002;i++){
if (t[i]>0&&g[i]) ans++;
}
cout<<ans<<endl;
return 0;
}

题目略坑，看懂就行

#include <stdio.h>
#include <string.h>
int main()
{
int n,ans;
int a[20000];
int b[20000];
memset(b,0,sizeof(b));
ans=0;
scanf("%d",&n);
for (int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
b[a[i]]=1;
}
for (int j=1;j<=n;j++)
{
for (int k=j+1;k<=n;k++)
{
if (b[a[j]+a[k]]==1)
{
b[a[j]+a[k]]=2;
}
}
}
for (int l=1;l<=n;l++)
{
if (b[a[l]]==2)
{
ans++;
}
}
printf("%d",ans);
return 0;
}

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

int main()
{
    int counter = 0;
    int num[100];
    int n;
    int i, j, k, flag;
    scanf("%d",&n);
    for(i = 0; i < n; i++)
        scanf("%d",num + i);
    for(i = 0; i < n; i++)
    {
        flag = 0;
        for(j = 0; j < n; j++)
            for(k = 0; k < n; k++)
                if(num[i] == (num[j] + num[k]) && i != k && i != j && j != k)
                    flag = 1;
        if(flag)
            counter++;
    }
    printf("%d",counter);
    return 0;
}

#include<iostream>
using namespace std;
int b[110];
int main(){
int n;
int a,sum=0;
cin>>n;
for(int i=0;i<n;i++){
cin>>a;
b[i]=a;
}
for(int i=0;i<n;i++)
{
int flag=0;
for(int j=0;j<n;j++)
{
for(int w=0;w<n;w++)
{
if(b[i]==b[w]+b[j]&&flag==0&&i!=w&&i!=j&&j!=w)
{
flag=1;
sum++;
}
}
}
}
cout<<sum;
return 0;
}

#include<stdio.h>
int main ()
{
int i,j,k=0,l=0,n,x,m=0;
scanf("%d",&n);
int a[n];
for(i=0;i<n;i++)
scanf("%d",&a[i]);

while(k<n)
{

for(i=0;i<n;i++)
for(j=i+1;j<n;j++)
if(a[i]+a[j]==a[k]&&a[i]!=a[j]&&k<n)
{
m++;
k++;
break;
}

if(l==k)
k++;
l++;

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

    int n, count = 0;
    scanf("%d",&n);
    int *a = (int *)malloc(sizeof(int)*n);
    int *c = (int *)malloc(sizeof(int)*20000);
    for (int i = 0; i != n;++i) {
    
        scanf("%d",a+i);
    
    }

    for (int i = 0; i != n - 1;++i) {
    
        for (int j = i + 1; j != n;++j) {
        
            c[a[i] + a[j]] = 1;
        
        }
    
    }

    for (int i = 0; i != n;++i) {
    
        if (c[a[i]]==1){
            ++count;
        }
    
    }
    printf("%d\n", count);
    free(a);
    free(c);