题解

171 条题解

  • 0
    @ 2017-08-14 10:57:22

    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;
    }
    
  • 0
    @ 2017-08-12 16:34:19

    #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;
    }

  • 0
    @ 2017-08-12 16:34:17

    #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;
    }

  • 0
    @ 2017-08-12 11:19:00

    #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;

    }

  • 0
    @ 2017-08-12 10:04:16

    #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;
    }

  • 0
    @ 2017-08-01 09:46:10

    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.

  • 0
    @ 2017-07-20 17:58:05

    #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;
    }

  • 0
    @ 2017-07-10 10:43:34

    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.
    
  • 0
    @ 2017-06-17 13:02:31

    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.

  • 0
    @ 2017-06-07 23:05:41

    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;
    }
    
    
  • 0
    @ 2017-05-25 16:47:25

    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;
    }
    
  • 0
    @ 2017-05-09 12:55:38
    #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);
    }
    
  • 0
    @ 2017-04-19 19:36:52

    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;
    }
    
  • 0
    @ 2017-04-15 09:45:35

    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);
    }
    
  • 0
    @ 2017-01-03 01:04:53

    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 ;
    }
    
    
  • 0
    @ 2016-12-22 18:59:57

    分站
    给洛谷投食
    应用

    搜索
    Alextokc
    P2141 珠心算测验
    1.6K
    通过
    5K
    提交
    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;
    }

  • 0
    @ 2016-11-25 21:55:04

    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;
    }

  • 0
    @ 2016-11-21 01:12:55
    #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;
    }
    
    
  • 0
    @ 2016-11-17 17:19:38

    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;
    }

    • @ 2016-11-28 14:12:26

      a[i] <= 10000
      bool b[20001]才对

  • 0
    @ 2016-11-15 13:33:28

    #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;
    }

信息

ID
1911
难度
6
分类
(无)
标签
递交数
16919
已通过
4473
通过率
26%
被复制
26
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