Mr. B has recently discovered the grid named "spiral grid".Construct the grid like the following figure. (The grid is actually infinite. The figure is only a small part of it.) Considering traveling in it, you are free to any cell containing a c…

  POJ 3126  Prime Path Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16204   Accepted: 9153 Description The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change…

题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=592 解决以下问题后就方便用广搜解: 1.将数字坐标化,10000坐标为(0,0),这样就可以通过数字获得其坐标 2.通过一个坐标知道在这个位置上的数字是否为素数 #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<queue> us…

题目: The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.— It is a matter of security to change such things every now and then,…

题目描述 题目背景 题目名称是吸引你点进来的[你怎么知道的] 实际上该题还是很水的[有种不祥的预感..] 题目描述 区间质数个数 输入输出格式 输入格式: 一行两个整数 询问次数n,范围m接下来n行,每行两个整数 l,r 表示区间. 输出格式: 对于每次询问输出个数 t,如l或r∉[1,m]输出 Crossing the line 输入输出样例 输入样例: 2 5 1 3 2 6 输出样例: 2 Crossing the line 说明 数据范围和约定 对于20%的数据 1<=n<=10 1&…

Catch That Cow Time Limit: 2000 MS Memory Limit: 65536 KB 64-bit integer IO format: %I64d , %I64u Java class name: Main [Submit] [Status] [Discuss] Description Farmer John has been informed of the location of a fugitive cow and wants to catch her imm…

这个题我看了,都是推荐的神马双向广搜,难道这个深搜你们都木有发现?还是特意留个机会给我装逼? Open the Lock Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3014    Accepted Submission(s): 1323 Problem Description Now an emergent task for you…

题目链接:http://acm.hust.edu.cn/vjudge/contest/128683#problem/E 题目大意:给定一只含有0和1的地图,0代表可以走的格子,1代表不能走的格 子.之后过了M年,每年把一个格子变成1, 即每年会有一个格子不能走.问地 图上界和下界连通共多少年.如果到M年之后依然连通,则输出-1. 给定的年份数据范围是10^5, 图的大小是500*500,由于图是一直在改变的, 因此每次都需要进行变更操作.然而如果变更一次广搜一次明显会TLE.仔细思 考会发现,当…

1.gcd int gcd(int a,int b){ return b?gcd(b,a%b):a; } 2.扩展gcd )extend great common divisor ll exgcd(ll l,ll r,ll &x,ll &y) { if(r==0){x=1;y=0;return l;} else { ll d=exgcd(r,l%r,y,x); y-=l/r*x; return d; } } 3.求a关于m的乘法逆元 ll mod_inverse(ll a,ll m){ l…

Codeforces Round #257 (Div. 1) C Codeforces Round #257 (Div. 1) E CF450E C. Jzzhu and Apples time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has picked n apples from his big apple tre…