北邮oj题库刷题计划（更新ing）

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 
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
#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");
        else printf("No\n");
    }
    return 0;
}

120 日期

#include "stdio.h"

int main(){
	int T;
	scanf("%d",&T);	//轮数
	while(T--){	//我发现这种写法还挺好用的 
		int y,m,d;
		int n=0;
		scanf("%d:%d:%d",&y,&m,&d);
		if(m==1) n=d;
		if(m==2) n=31+d;
		if(m>2){
			n=31+28;
			for(int i=3;i<m;i++){
				if((i<8&&i%2==1)||(i>7&&i%2==0)){
					n+=31;
				}
				else {
					n+=30;
				}
			}
			n+=d;
			if((y%100!=0&&y%4==0)||(y%100==0&&y%400==0)){
				n++;
			}
		}
		printf("%d\n",n);
	} 
	return 0;
} 

121 最值问题

#include
#include
using namespace std;

int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		int N;
		scanf("%d",&N);
		int a[N];
		for(int i=0;i<N;i++){
			scanf("%d",&a[i]);
		}
		sort(a,a+N);
		printf("%d %d\n",a[N-1],a[N-2]);
	}
}

122 统计时间间隔

#include 
using namespace std;

int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		int t1[2];
		int t2[2];
		scanf("%d:%d",&t1[0],&t1[1]);
		scanf("%d:%d",&t2[0],&t2[1]);
		int x = t1[0]*60+t1[1];
		int y = t2[0]*60+t2[1];
		int base = 24*60;
		printf("%d\n",(y-x+base)%base);	
	}
	return 0;
}

123 字符串转换

#include 
#include 
#include 
#include 
using namespace std;

//从a到z试中间数比较 

int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		char str[1001];
		int num[1001];
		cin>>str;
		for(int i=0;str[i]!='\0';i++){
			num[i] = (int) str[i]-'a';
		}
		int cost=10000;
		int k=0;
		for(int i=0;i<26;i++){
			int sum=0;
			for(int j=0;str[j]!='\0';j++){
				int x = abs(num[j]-i);
				if(x>13) sum += 26-x;
				else sum += x;
			}
			if(sum<cost) {
				cost = sum;
			}
		}
		printf("%d\n",cost);
	}	
	return 0;
}

124 文件系统

//文件系统，使用静态链表
#include 
using namespace std;
const int maxn=1010;
#define DIR  0
#define FILE  1
struct node{
	int type;
	string name;
	int parent; 
}Node[maxn];
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		//每轮都要初始化
		int index = 0;
		for(int i=0;i<maxn;i++){
			Node[i].parent = -1;
		}
		Node[0].name = "root";
		Node[0].type = FILE;
		int n;
		scanf("%d",&n);
		while(n--){
			string name1,name2;
			string op;
			cin>>op;
			if(op=="CREATEFILE"){
				cin>>name1>>name2;
				int i;
				for(i=0;i<=index;i++){
					if(name2==Node[i].name){
						index++;
						Node[index].name = name1;
						Node[index].type = FILE;
						Node[index].parent = i;
					}
				}
			}
			if(op=="CREATEDIR"){
				cin>>name1>>name2;
				int i;
				for(i=0;i<=index;i++){
					if(name2==Node[i].name){
						index++;
						Node[index].name = name1;
						Node[index].type = DIR;
						Node[index].parent = i;
					}
				}
			}
			string name;
			if(op=="LISTFILE"){
				cin>>name;
				for(int i=0;i<=index;i++){
					if(name==Node[i].name){
						for(int j=i;j<=index;j++){
							if(Node[j].parent == i&&Node[j].type==FILE) cout<<Node[j].name<<"\n";
						}
					}
				}
			}
			if(op=="LISTDIR"){
				cin>>name;
				for(int i=0;i<=index;i++){
					if(name==Node[i].name){
						for(int j=i;j<=index;j++){
							if(Node[j].parent == i&&Node[j].type==DIR) cout<<Node[j].name<<"\n";
						}
					}
				}
			}
		}
	}
	return 0;
} 

125 统计节点个数

#include "bits/stdc++.h"
using namespace std;
const int maxn = 1010;
struct Node{
	int du;
	int parent;
}node[maxn]; 
int main(){
	int T;
	cin>>T;
	while(T--){
		for(int i=0;i<maxn;i++){
			node[i].du = 0;
			node[i].parent = -1;
		}
		int N,n;
		cin>>n;
		N=n-1;
		while(N--){
			int x,y;
			cin>>x>>y;
			node[x].du++;
			node[y].du++;
			node[y].parent = x;
		}
		int num=0;
		for(int i=0;i<n;i++){
			//怎么也不对后来发现题目理解错了，不是大于等于所有儿子和父亲节点的度的和而是都不小于即可 
			if(node[i].parent==-1||node[node[i].parent].du<=node[i].du) {
				int j=0;
				while(j<n){
					if(node[j].parent==i){
						if(node[j].du>node[i].du) break;
					} 
					j++;
				}
				if(j==n) num++;
			}
		}
		cout<<num<<"\n";
	}
	return 0;
}

128 二进制数

#include

using namespace std;

int main()
{
	int T;
    scanf("%d",&T);
    while(T--){
    	unsigned int x;
    	scanf("%ud",&x);
    	stack<char> b;
    	while(x){
    		if(x%2) b.push('1'); 
    		else b.push('0');
    		x /= 2;
		} 
		while(!b.empty()){
			printf("%c",b.top());
			b.pop();
		}
		printf("\n");
		 
    }
    return 0;
}

299 分数加法

#include

using namespace std;

int main()
{
	int T;
    scanf("%d",&T);
    while(T--){
    	int a,b;
        scanf("%d %d",&a,&b);
        if(a<b){
        	swap(a,b);
		}
		int x = a-b;
		int m = pow(2,x)+1;	//分子 
		int n = pow(2,a);
		//注意处理a=b的情况
		if(m%2==0){
			m/=2;
			n/=2;
		} 
		printf("%d/%d\n",m,n);
    }
    return 0;
}

129 矩阵幂

#include

using namespace std;

int main()
{
	int T;
    scanf("%d",&T);
    while(T--){
    	int n,k;
    	scanf("%d %d",&n,&k);
    	int a[n][n],b[n][n],c[n][n];
    	for(int i=0;i<n;i++){
    		for(int j=0;j<n;j++){
    			scanf("%d",&a[i][j]);
    			b[i][j]=a[i][j];
			}
		}
		k--;
		while(k--){
			for(int i=0;i<n;i++){
				for(int j=0;j<n;j++)
				c[i][j]=0;
			}
			for(int i=0;i<n;i++){
    			for(int j=0;j<n;j++)
    				for(int m=0;m<n;m++)
    				c[i][j] += a[i][m]*b[m][j];
			}
			for(int i=0;i<n;i++){
				for(int j=0;j<n;j++)
				a[i][j]=c[i][j];
			}
		}
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				if(j==0) printf("%d",a[i][j]);	//保证输出格式 ,k=1的时候要输出a矩阵 
				else printf(" %d",a[i][j]);
			}
			printf("\n");
		}
    }
    return 0;
}