北郵oj題庫刷題計劃（更新ing）
83. A + B Problem
Calculate the sum of two given integers A and B.
輸入格式
The input consists of a line with A and B. (−104≤A,B≤104).
輸出格式
Output the only answer.
輸入樣例
2 3
輸出樣例
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.
輸入格式
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].
輸出格式
Output the answer for each test case, one per line.
輸入樣例
4
1 1 1 3
10
1 2 3 1 2 3 1 2 3 4
輸出樣例
3
4
#include <bits/stdc++.h>
using namespace std;
map<unsigned long long int,int> input; //一定注意範圍！ 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
題目描述
Given points on a 2D plane, judge whether there’re three points that locate on the same line.
輸入格式
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].
輸出格式
For each test case, output Yes if there’re three points located on the same line, otherwise output No.
//怎麼說呢，我真是菜雞，一開始以爲是精度的問題...但實際上去掉eps也是對的...我瘋了，還有要注意斜率爲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");
else printf("No\n");
}
return 0;
}
120 日期
#include "stdio.h"
int main(){
int T;
scanf("%d",&T); //輪數
while(T--){ //我發現這種寫法還挺好用的
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 最值問題
#include<cstdio>
#include<algorithm>
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 統計時間間隔
#include <cstdio>
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 字符串轉換
#include <cstdio>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
//從a到z試中間數比較
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 文件系統
//文件系統，使用靜態鏈表
#include<bits/stdc++.h>
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--){
//每輪都要初始化
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 統計節點個數
#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++){
//怎麼也不對後來發現題目理解錯了，不是大於等於所有兒子和父親節點的度的和而是都不小於即可
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 二進制數
#include<bits/stdc++.h>
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 分數加法
#include<bits/stdc++.h>
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);
//注意處理a=b的情況
if(m%2==0){
m/=2;
n/=2;
}
printf("%d/%d\n",m,n);
}
return 0;
}
129 矩陣冪
#include<bits/stdc++.h>
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]); //保證輸出格式 ,k=1的時候要輸出a矩陣
else printf(" %d",a[i][j]);
}
printf("\n");
}
}
return 0;
}