北邮oj题库刷题计划（更新ing）

最新推荐文章于 2021-05-18 02:55:52 发布

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最新推荐文章于 2021-05-18 02:55:52 发布

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分类专栏：北邮文章标签： oj系统 c语言

本文链接：https://blog.youkuaiyun.com/weixin_39938635/article/details/103996577

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这篇博客介绍了在北邮在线判题系统（OJ）中的一系列编程题目，包括A + B Problem、Single Number和Three Points On A Line等。内容涵盖数字运算、数组处理、几何问题等多个方面，适合C语言学习者进行编程练习和算法提升。

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83. A + B Problem

Calculate the sum of two given integers A and B.

输入格式
The input consists of a line with A and B. (−104≤A,B≤104).

输出格式
Output the only answer.

输入样例
2 3
输出样例
5

#include "stdio.h"
 
int main(){
   
    int a,b = 0;
    scanf("%d %d", &a,&b);
    int c = a+b;
    printf("%d",c);
    return 0;
}

84 Single Number

Given an array with N integers where all elements appear three times except for one. Find out the one which appears only once.

输入格式
Several test cases are given, terminated by EOF.

Each test case consists of two lines. The first line gives the length of array N(1≤N≤105), and the other line describes the N elements. All elements are ranged in [0,263−1].

输出格式
Output the answer for each test case, one per line.

输入样例
4
1 1 1 3
10
1 2 3 1 2 3 1 2 3 4
输出样例
3
4

#include <bits/stdc++.h>
using namespace std;
map<unsigned long long int,int> input;	//一定注意范围！ WA了好多遍 

int main(){
   
	int n;
	while(scanf("%d",&n)!=EOF){
   
		input.clear();
		unsigned long long int a[n];
		for(int i=0;i<n;i++){
   
			scanf("%llu",&a[i]);
			if(input.find(a[i])==input.end()) input[a[i]]=1;
			else input[a[i]]++;
		}
		for(int i=0;i<n;i++){
   
			if(input[a[i]]==1)
			{
   
				printf("%llu\n",a[i]);
				break;
			}
		}

		
	}
	return 0;
}

85. Three Points On A Line

题目描述
Given points on a 2D plane, judge whether there’re three points that locate on the same line.

输入格式
The number of test cases T(1≤T≤10) appears in the first line of input.

Each test case begins with the number of points N(1≤N≤100). The following N lines describe the coordinates (xi,yi) of each point, in accuracy of at most 3 decimals. Coordinates are ranged in [−104,104].

输出格式
For each test case, output Yes if there’re three points located on the same line, otherwise output No.

//怎么说呢，我真是菜鸡，一开始以为是精度的问题...但实际上去掉eps也是对的...我疯了，还有要注意斜率为0的，要用乘的形式...
#include<bits/stdc++.h>
#define MAXN 105

int main()
{
   
	int T, i, j, k, m, n, flag;
	double a[MAXN], b[MAXN];
    scanf("%d",&T);
    while(T--){
   
        flag=0;
        scanf("%d",&n);
        scanf("%lf %lf",&a[0],&b[0]);
        scanf("%lf %lf",&a[1],&b[1]);
        for(int m=2;m<n;m++){
   
            scanf("%lf %lf",&a[m],&b[m]);
            for(i=0;i<m-1&&!flag;i++) 
				for(int j=i+1;j<m;j++)
                if(fabs(a[i]*b[j]-a[j]*b[i]+a[j]*b[m]-a[m]*b[j]+a[m]*b[i]-a[i]*b[m])<=1e-6) flag=1;
        }
        if(flag) printf("Yes\n");</

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