记录一下一个业务问题，流程是这样的，我现在有一个定时任务，5分钟执行一次，更新车辆打卡的情况。现在有20俩车，每辆车都分配了路线，每条路线都有打卡点，每个打卡点分配了不同的时间段，也就是说，一条路线可能有几百个打卡点，这几百个打卡点中每一个都分配了时间段，有可能是1个时间段，比如8

3302.表达式求值-AcWing题库#include<bits/stdc++.h>usingnamespacestd;#defineintlonglong#defineendl'\n'constintN=5e5+10;stack<int>st;//存数字stack<char>op;//存操作符voideval(){inta=st.top();//第二个数st.pop();

task4#include<stdio.h>intfunc(){intans,i;charc;FILE*fp;fp=fopen("data4.txt","r");while(fp==NULL){printf("failtoopenfile\n");return0;}i=0;while(!feof(fp))

task4#include<stdio.h>intmain(){FILE*fp;charch;intcount=0;fp=fopen("d:\\data4.txt","r");if(fp==NULL){printf("failtoopenfile\n");return0;}while(!feof(fp)){

TASK4点击查看代码#include<stdio.h>intmain(){intcount=0;intch;FILE*fp;fp=fopen("data4.txt","r");if(fp==NULL){printf("failtoopenfile\n");return1;}

1.#include<stdio.h>#include<stdlib.h>intmain(){intch=0;inta=0;FILE*fp;fp=fopen("data4.txt","r");if(fp==NULL){printf("failtoopen\n");return1;}whil

任务一代码`#include<stdio.h>include<stdlib.h>defineN80defineM100typedefstruct{charname[N];//书名charauthor[N];//作者}Book;//函数声明voidfunc1();voidfunc2();intmain(){func1();func2();system("pause");return0;

思路因为此题目中对于数的更新只能为\(1\)，所以，如果我们找到了\([1,l-1]\)与\([r+1,n]\)中能获得的两个极值即可。我们为\(S\)赋予权值，用一个数组\(a\)储存：如果\(S_i\)为+，则其权值为\(1\)。否则其权值为\(-1\)。那么，在第\(i\)次操作后，能产生的数是\(\s

4.实验任务4：1#include<stdio.h>23intmain()4{5inti=0;6FILE*fp;7charstr[100];8charch;910fp=fopen("data4.txt","r");11if(fp==NULL)12{13printf("failt

TASK1#include<stdio.h>#include<Stdlib.h>#defineN80#defineM100typedefstruct{charname[N];charauthor[M];}Book;voidfunc1();voidfunc2();intmain(){func1();func2();system("pause");return0;

task1#include<stdio.h>#defineN80#defineM100typedefstruct{charname[N];//书名charauthor[N];//作者}Book;//函数声明voidfunc1();voidfunc2();intmain(){func1();func2();return0;}//函数func1定义

1#include<stdio.h>2#include<stdlib.h>3intmain()4{5intch=0;6inta=0;7FILEfp;89fp=fopen("data4.txt","r");1011if(fp==NULL){12printf("failtoopen\n");13r

4.task.4#include<stdio.h>intmain(){FILE*fp;charch;intcount=0;fp=fopen("d:\\data4.txt","r");if(fp==NULL){printf("failtoopenfile\n");return0;}while(!feof(fp))

1#include<stdio.h>2#include<stdlib.h>3intmain()4{5intch=0;6inta=0;7FILE*fp;89fp=fopen("data4.txt","r");1011if(fp==NULL){12printf("failtoopen\n");

#include<stdio.h>intmain(){intcnt=0;charch;FILE*fp;fp=fopen("data4.txt","r");if(fp==NULL){printf("failtoopenfile\n");return0;}while((ch=fgetc(fp))!=

`#include<stdio.h>include<stdlib.h>defineN80defineM7typedefstruct{charname[N];//书名charauthor[N];//作者}Book;//函数声明voidfunc1();voidfunc2();intmain(){func1();func2();system("pause");return0;}//函数fu

实验任务4#include<stdio.h>intmain(){FILE*fp;charch;intcount=0;fp=fopen("d:\\data4.txt","r");if(fp==NULL){printf("failtoopenfile\n");return0;}while(!feof(fp)){

task4#include<stdio.h>intmain(){intcnt=0;charch;FILE*fp;fp=fopen("data4.txt","r");if(fp==NULL){printf("failtoopenfile\n");return;}while((ch=fgetc(f

task1//将图书信息写入文本文件data1.txt//再从文件中读取图书信息，打印输出到屏幕上，并显示行号#include<stdio.h>#defineN80#defineM100typedefstruct{ charname[N];//书名 charauthor[N];//作者}Book;//函数声明voidfunc1();voidfunc2()

task1点击查看代码//将图书信息写入文本文件data1.txt//再从文件中读取图书信息，打印输出到屏幕上，并显示行号#include<stdio.h>#include<stdlib.h>#defineN80#defineM100typedefstruct{ charname[N];//书名 charauthor[N];//作者}Book;//

task4#include<stdio.h>intmain(){intcount;charch;FILE*fp;fp=fopen("data4.txt","r");if(fp==NULL){printf("failtoopenfile\n");return1;}count=0;wh

思路定义\(dp_{i,j}\)表示区间\([i,j]\)中回文串的数量。那么，不难得出状态转移方程\(dp_{i,j}=dp_{i-1}+f_{i,j}\)。（其中\(f_{i,j}\)表示左端点大于等于\(i\)，右端点为\(j\)的回文串数量）由此，现在问题转变为了如何求\(f_{i,j}\)。如果我们在求出了\(f_{i+1,j}

思路首先假设有两个串\(a,b\)，如果\(b\)是\(a\)的子串，且\(a\neqb\)则不需要考虑\(b\)；如果\(a=b\)，则如需要保留一个\(a\)。做完上述操作后，显然最终的答案是由这些串按照一定顺序拼接起来，再删掉重叠部分。例如：abbcc与ccdde拼接为abbccccdde，发现cc是重复的，所以

task4#include<stdio.h>intmain(){intcount=0;intch;FILE*fp;fp=fopen("data4.txt","r");if(fp==NULL){printf("failtoopenfile\n");return1;}while((