Reticent_Man的博客

题目描述There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations i...

The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters.Now ...

This time, you are supposed to find A+B where A and B are two polynomials.Input Specification:Each input file contains one test case. Each case occupies 2 lines, and each line contains the informa...

Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).Input Specification:Each input ...

#include <iostream>using namespace std;int main(){ int n,a[101]; cin>>n; for(int i=0;i<n;i++) { cin>>a[i]; } int sum=a[0]*6; for(int i=1;i<n;i++) { int t=a[i]-a[...

#include <iostream>#include <cstring>using namespace std;struct student{ char id[16]; char ftime[9]; char ltime[9];};int main(){ int n; cin>>n; student a[201]; for(int ...

#include <iostream>#include <cstring>using namespace std;int change(int m){ switch(m) { case 0:cout<<"zero"; break; case 1:cout<<"one"; break; case 2:cout<&a

#include <iostream>#include <cstring>#include <cmath>using namespace std;int main(){ int n,a[10010],b[10010],maxx=0; cin>>n; memset(a,0,sizeof(a)); memset(a,0,sizeof(...

There is a classical process named partition in the famous quick sort algorithm. In this process we typically choose one element as the pivot. Then the elements less than the pivot are moved to its le...

For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in non-increasing order first, and then in non-decreasing order, a new number can be obtained by takin...

题目描述Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the pa...

Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists w...

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×p​m​​​k​m​​​​.Input Specification:Each input file co...

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the l...

#include <iostream>using namespace std;int main(){ int n,a[10001],x[10001],y[10001],maxx=0,minn=30000,b,c; cin>>n; for(int i=0;i<n;i++) { cin>>a[i]>>x[i]>>y[i...

#include <iostream>#include <cstring>using namespace std;int main(){ int n; char a[81]; cin>>n; while(n--) { int x=0,y=0; memset(a,0,sizeof(a)); cin>>a; if(str...

#include <iostream>#include <cmath>using namespace std;int main(){ int m,n,sum=0,x=2; cin>>m>>n; while(sum!=n) { int ok=1; for(int i=2;i<=sqrt(x);i++) { if(...

#include <iostream>#include <iomanip>using namespace std;int main(){ int n,a[1001],a1=0,a2=0,a3=0,a4=0,a5=0,sum=0; cin>>n; int t=0,b[1001]; for(int i=0; i<n; i++) { c...

#include <iostream>using namespace std;int main(){ int T; long long a,b,c; cin>>T; for(int i=1;i<=T;i++) { cin>>a>>b>>c; cout<<"Case #"&

/*#include<cstdio>int main(){ int num = 0; //单词的个数 char ans[80][80]; while(scanf("%s", ans[num]) != EOF){ //一直输入直到文件末尾 num++; //单词的个数加 1 } for(int i = nu...

#include <iostream>#include <cstring>using namespace std;int main(){ int n,m,a[101]; cin>>n>>m; m=m%n; for(int i=0;i<n;i++) { cin>>a[i]; } int sum=0; for...

#include <iostream>#include <algorithm>using namespace std;int main(){ int n,a[200],b[900]={0},c[500]={0}; cin>>n; for(int i=0;i<n;i++) { cin>>a[i]; } sort(a,a...

#include <iostream>#include <cstring>using namespace std;int maxx=0,minn=9999999,cnt;int n,m,c1,c2,man[501],a[501],b[501],c[501][501],r[501];bool pan[501];int dfs(int from,int to,int...

#include <iostream>#include <iomanip>#include <cmath>#include <cstring>using namespace std;int main(){ int k1,k2; while(cin>>k1) { double a[1001]= {-1}; f...

#include <iostream>#include <iomanip>using namespace std;int main(){ long long a,a1,b,b1; cin>>a>>a1>>b>>b1; int sum1=0,sum2=0; while(a) { int t=a%10; ...

本题要求计算A/B，其中A是不超过1000位的正整数，B是1位正整数。你需要输出商数Q和余数R，使得A = B * Q + R成立。输入格式：输入在1行中依次给出A和B，中间以1空格分隔。输出格式：在1行中依次输出Q和R，中间以1空格分隔。输入样例：123456789050987654321 7输出样例：17636684150141093474 3

#include <iostream>#include <algorithm>#include <cstring>using namespace std;int main(){ long long n,p; int a[100001],maxx=0; cin>>n>>p; for(int i=0;i<n;i++) ...

#include <iostream>#include <cstring>#include <cmath>using namespace std;int main(){ int n; char a,b[1001][1001]={' '}; cin>>n; getchar(); cin>>a; int t=sqrt((...

#include <iostream>#include <cstring>using namespace std;int main(){ char a[1000]; cin>>a; int flag,flag1; for(int i=1; i<strlen(a); i++) { if(a[i]=='E') flag=i; }...

#include <iostream>using namespace std;int main(){ int a[10]; for(int i=0;i<10;i++) { cin>>a[i]; } for(int i=1;i<=9;i++) { if(a[i]) { cout<<i; a[i]--; ...

#include <iostream>using namespace std;int main(){ unsigned long long a,b,c; int d; cin>>a>>b>>d; c=a+b; if(c==0) cout<<"0"; int m[100]; int x=0; while(c) { ...

#include <iostream>#include <cstring>using namespace std;int main(){ char a[1001]; cin>>a; int b[1002]={0}; for(int i=0;i<strlen(a);i++) { int t=a[i]-'0'; b[t]++; }...

#include <iostream>#include <algorithm>#include <iomanip>#include <cstring>using namespace std;int main(){ int n,d; cin>>n>>d; double a[1001],b[1001],c[100...

#include <iostream>#include <cstring>#include <algorithm>using namespace std;int main(){ int n,t1=0,t2=0,x=0; cin>>n; while(n!=6174||x==0) { x++; t1=0,t2=0; ...

#include <iostream>#include <cmath>#include <cstring>using namespace std;int main(){ int a,b,c,m[10]= {0},x=0; cin>>a>>b; c=a+b; int ok=0; if(c>=0) ok=1; el...