qq_43789349的博客

#include<iostream>#include<queue>using namespace std;struct Node{ int lchild; int rchild;}node[110];bool isnotroot[110]={false};int n,num=0;void print(int id){ cout<<id; num++; if(num!=n) cout<&

#include<iostream>#include<stack>#include<queue>using namespace std;struct node{ int data; node* lchild; node* rchild;};int pre[50],in[50];int n;node* create(int prel,int prer,int inl,int inr){ if(prel>prer

#include<iostream>#include<vector>#include<algorithm>using namespace std;const int maxn=110;struct Node{ int weight; vector<int> child;}node[maxn];bool cmp(int a,int b){ return node[a].weight>node[b].weight

#include<iostream>#include<queue>#include<algorithm>#include<stdio.h>using namespace std;struct node{ int data; node* lchild; node* rchild;};const int maxn=50;int n,pre[maxn],post[maxn],in[maxn];node* create(

首先讲一下思路。这道题是一个贪心算法的题，大体方法就是0.寻找比自己距离远的，到能够到达的最大距离之间的加油站，看他们的油价。如果找到了更低价格的油价，就加油到刚好能到达那个加油站的距离的油，然后去那个更低价格的加油站。如果找不到更低的，就找尽可能低的油价的加油站，在当前加油站加满油之后过去。因为想要让路程上使用的尽可能是低价的油，既然没有比当前更低价格的了，就让油箱加到最大值，这样能保证利益最大化，保证最大的距离使用的是便宜的油。但有几个要注意的地方。第一个是终点的处理。基本所有的解答都是将终点当作

#include<iostream>#include<cmath>#include<map>#include<set>using namespace std;bool isprime(long long num){ for(long long i=2;i<=sqrt(num);i++) if(num%i==0) return false; return true;}int main(){ lo

#include<iostream>#include<string>#include<cstring>using namespace std;int main(){ string s1,s2; cin>>s1; int epoint; for(int i=0;i<s1.length();i++) { if(s1[i]=='E') { epoint=

题目大意：给出两个字符串，在第一个字符串中删除第二个字符串中出现过的所有字符并输出分析：用flag[256]数组变量标记str2出现过的字符为true，输出str1的时候根据flag[str1[i]]是否为true,如果是true就不输出注意：使用int lens1 = strlen(s1);int lens2 = strlen(s2);的形式，否则直接放在for循环里面会超时#include<iostream>#include<string>#include<cst

#include<iostream>#include<algorithm>using namespace std;struct Student{ string id; string name; int score;};Student student[10020];bool cmp1(Student a,Student b){ return a.id<b.id;}bool cmp2(Student a,Stude

#include<iostream>#include<cmath>using namespace std;int isprime(int n){ if(n==1) return -1; else { for(int i=2;i<=sqrt(n);i++) if(n%i==0) return -1; return 1; }}int main(){

最大公约数long long gcd(long long a,long long b){ return !b?a:gcd(b,a%b);}最小公倍数long long findmin(long long a,long long b){ long long temp=gcd(a,b); return a*b/temp;}

#include<iostream>using namespace std;long long gcd(long long a,long long b){ return !b?a:gcd(b,a%b);}long long findmin(long long a,long long b){ long long temp=gcd(a,b); return a*b/temp;}int main(){ long long n; long

#include<iostream>#include<cmath>#include<algorithm>#include<iomanip>using namespace std;struct Mooncake{ float demand; float price; float xjb;};Mooncake mooncake[1005];bool cmp1(Mooncake a,Mooncake b){ re

#include<iostream>#include<string>#include<cstring>#include<algorithm>#include<bits/stdc++.h>using namespace std;struct Student{ string name; string gender; //这里写string 写char都可以 string id; int grade;}st

#include<iostream>#include<string>#include<vector>using namespace std;int main(){ string s; cin>>s; int n1,n2,n3,n; n=s.length()+2; if(n%3==0) n2=n/3; else if(n%3==1) n2=n/3+1; else

#include#include#includeusing namespace std;void change(int n){int m=0;vector temp;do{temp.push_back(n%13);n=n/13;}while(n!=0);char c=‘A’;if(temp.size()==1){cout<<‘0’;if(temp[0]>9)printf("%c",‘A’+temp[0]-10);elsecout<<temp

#include#include#includeusing namespace std;struct client{string name;int month;int day;int hour;int minute;string statue;}per[1000];int fee[24];bool cmp1(client a,client b){if(a.name!=b.name)return a.name<b.name;else if(a.month!=b.mon

#include<iostream>#include<algorithm>using namespace std;struct node{ int address; int key; int next; int exist;}nodes[100005];bool cmp1(node a,node b){ if(a.exist==0||b.exist==0) return a.exist>b.exist;

#include<iostream>#include<string>#include<map>#include<vector>using namespace std;char word[100005];int node[100005];int main(){ int firadd1,firadd2,n; cin>>firadd1>>firadd2>>n; int add,next;

#include<iostream>#include<string>#include<algorithm>using namespace std;struct people{ string name; int talent; int virtue;};bool cmp1(people a,people b){ int totala=a.virtue+a.talent; int totalb=b.virtue+b.t

#include#include#includeusing namespace std;int main(){int n;cin>>n;string a,b;cin>>a>>b;vector suma;vector sumb;int kinda=0;int kindb=0;int flaga=0,flagb=0;if(a.find(".")!=string::npos){if(a[a.find(".")-1]==‘0’&&

#include<iostream>#include<algorithm>using namespace std;struct student{ int teamno; string name; int score; int scoreno; //小组排名 int totalno; //总分排名}stu[30000];bool cmp1(student a,student b){ if(a.score!=b.score)

A reversible prime in any number system is a prime whose “reverse” in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.Now given any two positive integers N (<105) a

To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algrbra), and E - English. At the mean time, we encourage students by emphasi

1011 World Cup Betting (20 分)With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excited as the best players from the best teams doing battles for the World Cup trophy in South Africa. Similarly, football betting

1096 Consecutive Factors (20 分)Among all the factors of a positive integer N, there may exist several consecutive numbers. For example, 630 can be factored as 3×5×6×7, where 5, 6, and 7 are the three consecutive numbers. Now given any positive N, you are

仅22分 测试点0,7,17不能通过即上下限，大数不能通过，懒得用二分法1010 Radix (25 分)Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is yes, if 6 is a decimal number and 110 is a binary number.Now for any pair of positive integ

1007 Maximum Subsequence Sum (25 分)Given a sequence of K integers { N1​, N2​, …, NK​}. A continuous subsequence is defined to be { Ni​, Ni+1​, …, Nj​} where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the lar

1008 Elevator (20 分)The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one fl

Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.Input Specification:Each input file contains one test case. Each case occupies one line which contains an N (≤10100).O

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 length of each road between any pair of cities are marked

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 file contains one test case. Each case contains a pair of i

This time, you are supposed to find A+B where A and B are two polynomials.InputEach input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 … NK aNK, where K is the number of

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).InputEach input file contains one test case. Each case contains a pair of integers a and