题解

171 条题解

  • 0
    @ 2021-05-17 15:32:43

    #include <iostream>
    using namespace std;
    bool flag[20001]={false};
    int main()
    {
    int n,a[101],count=0;
    cin>>n;
    for(int i=1;i<=n;i++){
    cin>>a[i];
    for(int j=1;j<i;j++){
    flag[a[i]+a[j]]=true;
    }
    }
    for(int i=1;i<=n;i++){
    if(flag[a[i]]){
    count++;
    }
    }
    cout<<count<<endl;
    }

  • 0
    @ 2021-01-20 21:44:19

    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)
            
    
    
  • 0
    @ 2020-12-22 21:41:39
    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)
    
  • 0
    @ 2020-05-19 05:32:41

    #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);
    }

  • 0
    @ 2020-04-13 14:08:33
    #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;
    }
    
    
  • 0
    @ 2020-03-10 18:02:27

    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);
    }
    
  • 0
    @ 2020-03-05 22:18:26

    #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;
    }

  • 0
    @ 2019-12-22 10:30:55

    #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;
    }

  • 0
    @ 2019-12-12 14:31:10

    #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;
    }

  • 0
    @ 2019-11-21 09:17:35

    #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;
    }

  • 0
    @ 2019-11-15 15:24:35

    /*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;
    }
    
  • 0
    @ 2019-10-19 23:58:14

    #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;
    }

  • 0
    @ 2018-11-04 12:52:40

    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;
    }

  • 0
    @ 2018-10-07 08:26:04

    #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;
    }

  • 0
    @ 2018-10-06 11:02:01

    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;
    }

  • 0
    @ 2018-08-16 20:34:44
    #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;
    }
    
    
  • 0
    @ 2018-08-07 15:03:49

    #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;
    }

  • 0
    @ 2018-07-19 20:26:48

    #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;
    }

  • 0
    @ 2018-04-25 19:54:39
        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);
    
  • 0
    @ 2018-03-18 20:24:07

    数据太水了,暴力都能过……

    #include<bits/stdc++.h>
    using namespace std ;
    //Vijos P1911
    
    bool x[20000] ;
    int main()
    {
        int y[100] , n , ans = 0 ;
        scanf("%d" , &n) ;
        for (int i = 0 ; i < n ; i ++)
        {
            scanf("%d" , &y[i]) ;
        }
        for (int i = 0 ; i < n - 1 ; i ++)
            for (int j = i + 1 ; j < n ; j ++)
            {
                x[y[i] + y[j]] = true ;//有这个和的打上标记
            }
        for (int i = 0 ; i < n ; i ++)
        {
            if (x[y[i]])
                ans ++ ;//打过标记的个数就是答案
        }
        printf("%d" , ans) ;
        return 0 ;
    }
    

信息

ID
1911
难度
6
分类
(无)
标签
递交数
16923
已通过
4475
通过率
26%
被复制
26
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