u013268685的专栏(有一种幸福叫AC，有一种期待叫AK)

虽然不难，但感觉挺不错，值得一做的字符串题

City GameTime Limit: 3000msMemory Limit: 131072KBThis problem will be judged on UVALive. Original ID: 302964-bit integer IO format: %lld      Java class name: MainPrev Submit Statu

好久没敲快排了，今天搞了半天居然不会，先存档再说，rtvoid quicksort(int *A, int low, int high){ int m=A[low]; int r=high,l=low; while(l<r) { while(A[l]<m)l++; while(A[r]>m)r--; if(l<=r){ int t=A[l];

瑶瑶的第K大Time Limit: 10000/5000MS (Java/Others)Memory Limit: 512000/256000KB (Java/Others)SubmitStatisticNext ProblemProblem Description一天，萌萌的妹子--瑶瑶(tsyao)很无聊，就来找你玩。可是你们都不知道玩什么。。。尴尬了一阵子

Description  Team Queue Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though

A - LIGHTOJ 1005Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %lluSubmit StatusDescriptionA rook is a piece used in the game of chess which is played on a board

应该是概率的题，还有最大公约数约分的问题。E - LIGHTOJ 1027Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %lluSubmit StatusDescriptionYou are in a maze; seeing n doors in fr

#include #include #include #define max 10001int a[max],b[max];void search(int num,int N){ int low=1,high=N,mid; int ok=0; mid=(low+high)/2; while(low<high) { if(a[mid

本题注意要查找一下输入的人的国家是否已经出现Problem BList of ConquestsInput: standard inputOutput: standard outputTime Limit: 2 secondsIn Act I, Leporello is telling Donna Elvira about his master's long list

字符串处理题，需要注意的是要一直替换到没有所给的字符串为止，所以可以在change函数里返回1（代表查找成功）和0（代表没有要替换的字符了）Problem E: Automatic EditingSource file:autoedit.{c, cpp, java, pas}Input file:autoedit.inOutput fil

本题要注意测试数据相当的多啊，所以要先构造出所有解并排序，然后用二分查找即可Bob's ProblemAccepted : 29 Submit : 193Time Limit : 1000 MS Memory Limit : 65536 KB 题目描述Bob今天碰到一个问题，他想

A Love LetterAccepted : 33 Submit : 66Time Limit : 1000 MS Memory Limit : 65536 KB 题目描述　　CodeMonkey终于下定决心用情书的方式向心爱的女神表白，当他历经几天几夜写完之后才知道女神有很多不喜欢的词，所以他不得不有把

C.长长长长龙Time Limit: 3000 MSMemory Limit: 32768 KTotal Submit: 202 (73 users)Total Accepted: 50 (46 users)Special Judge: NoDescription时间：今天是

需要注意对小数点的处理#include#include#include#include#define MAX 600#define LL strlenvoid jia(const char *str1,char *str)//add str1 to str{ int len1=LL(str1); int c=0;len1--; for(int i=MAX

要说的都在注释里#include#include#define MAX 10010#define LL strlenchar INT_MAX[15]="2147483648";int judge(char *str){ int len=LL(str),ok=LL(INT_MAX); if(len<ok)return 0; if(len==ok&&strcmp(INT_

#include#define Max 300#includechar str1[Max],str2[Max];int s[2*Max],c[2*Max];void jia(int *a,int n){ int c=0,ans=0; for(int i=0;i<=n;i++) { s[ans]+=a[i]+c; c=s[ans]/10

Distant GalaxyTime Limit: 3000msMemory Limit: 131072KBThis problem will be judged on UVALive. Original ID: 369564-bit integer IO format: %lld      Java class name: MainPrev Submit