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P2141 [NOIP2014 Popularization group ] Abacus mental arithmetic test

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The time limit 1.00s

Memory limit 125.00MB

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Title Description

Abacus mental calculation is a kind of calculation technology that can complete fast calculation by simulating the change of abacus in the brain . Abacus mental arithmetic training , Can develop intelligence , It can bring a lot of convenience to our daily life , So it's popularized in many schools .

Some school's abacus mental arithmetic teacher uses a kind of quick examination abacus mental arithmetic addition ability test method . He randomly generates a set of positive integers , The numbers in the set are different , Then ask the students to answer ： How many of them , Exactly equal to the other two in the set （ Different ） Sum of the numbers ？

Recently, the teacher gave some test questions , Please help me find out .

( This topic is 2014NOIP Universal T1)

Input format

There are two lines , The first line contains an integer nn, The number of positive integers given in the test .

The second line has nn A positive integer , Every two positive integers are separated by a space , A positive integer given in a test .

Output format

An integer , The answer to a test question .

I/o sample

Input #1 Copy

4
1 2 3 4

Output #1 Copy

2

explain / Tips

【 Sample explanation 】

from 1+2=3,1+3=41+2=3,1+3=4, So the answer to meet the test requirements is 22.

Be careful , The addend and the addend must be two different numbers in the set .

【 Data description 】

about 100\%100% The data of ,3 ≤ n ≤ 1003≤n≤100, The size of the positive integer given by the test question does not exceed 10,00010,000.

There is a hole in this question

such as 2+3=5 , and 1+4=5 It's a kind of , Therefore, it is necessary to check the duplicate , So I can mark

#include<cstdio>
#include<iostream>
using namespace std;
int a[105]={0};
int main(){
    int n=0;
    scanf("%d",&n);
    
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    int x=0,y=0,all=0;
    for(int i=1;i<n;i++)
    {
        x=a[i];
        for(int j=i+1;j<=n;j++)
        {
            y=a[j];
            for(int h=1;h<=n;h++)
            {
                if(x+y==a[h])
                {
                    all++;
                }
            }
        }
    }
    printf("%d",all);
    return 0;
} No duplicate check code

According to the code of the topic

#include<cstdio>
#include<iostream>
using namespace std;
int a[105]={0},b[105];
int main(){
    int n=0;
    scanf("%d",&n);
    
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        b[i]=2;
    }
    int x=0,y=0,all=0;
    for(int i=1;i<n;i++)
    {
        x=a[i];
        for(int j=i+1;j<=n;j++)
        {
            y=a[j];
            for(int h=1;h<=n;h++)
            {
                if(x+y==a[h]&&b[h]!=1)
                {
                    all++;
                    b[h]=1;
                }
            }
        }
    }
    printf("%d",all);
    return 0;
}