题目连接 http://poj.org/problem?id=2046 Gap Description Let's play a card game called Gap. You have 28 cards labeled with two-digit numbers. The first digit (from 1 to 4) represents the suit of the card, and the second digit (from 1 to 7) represents the…

题目地址: http://poj.org/problem?id=2046 一道搜索状态压缩的题目,关键是怎样hash. AC代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include <vector> #include <list&…

Description Let's play a card game called Gap. You have cards labeled with two-digit numbers. The first digit ( to ) represents the suit of the card, and the second digit ( to ) represents the value of the card. First, you shu2e the cards and lay the…

Power Strings Time Limit: 3000 MS Memory Limit: 65536 KB 64-bit integer IO format: %I64d , %I64u Java class name: Main [Submit] [Status] [Discuss] Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc…

OJ上的一些水题(可用来练手和增加自信) (POJ 3299,POJ 2159,POJ 2739,POJ 1083,POJ 2262,POJ 1503,POJ 3006,POJ 2255,POJ 3094) 初期: 一.基本算法: 枚举. (POJ 1753,POJ 2965) 贪心(POJ 1328,POJ 2109,POJ 2586) 递归和分治法. 递推. 构造法.(POJ 3295) 模拟法.(POJ 1068,POJ 2632,POJ 1573,POJ 2993,POJ 2996) 二…

著名题单,最初来源不详.直接来源:http://blog.csdn.net/a1dark/article/details/11714009 OJ上的一些水题(可用来练手和增加自信) (POJ 3299,POJ 2159,POJ 2739,POJ 1083,POJ 2262,POJ 1503,POJ 3006,POJ 2255,POJ 3094) 初期: 一.基本算法: 枚举. (POJ 1753,POJ 2965) 贪心(POJ 1328,POJ 2109,POJ 2586) 递归和分治法. 递…

POJ 3518 Prime Gap(素数) id=3518">http://poj.org/problem? id=3518 题意: 给你一个数.假设该数是素数就输出0. 否则输出比这个数大的素数与比这个数小的素数的差值. 分析: 明显本题先要用筛选法求出130W(个素数)以内的全部素数. 然后推断给的数是否是素数就可以. 假设不是素数.那么就找出它在素数素组内的上界和下界,输出两个素数的差值就可以. 筛选法求素数可见: http://blog.csdn.net/u013480600/a…

[题意简述]:输入一个数,假设这个数是素数就输出0,假设不是素数就输出离它近期的两个素数的差值,叫做Prime Gap. [分析]:这题过得非常险.由于我是打的素数表. 由于最大的素数是1299709,所以注意在打表时要使用long long.否则程序应该不能执行.注意这一点应该就能够了. 积累! // 2984K 235Ms #include<iostream> using namespace std; #define N 2000001 bool isprime[N]; long long…

Prime Gap Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 7392   Accepted: 4291 Description The sequence of n − 1 consecutive composite numbers (positive integers that are not prime and not equal to 1) lying between two successive prime…

/*以核心1为源点,以核心2为汇点建图,跑一遍最大流*/ #include<stdio.h> #include<string.h> #include<queue> using namespace std; #define N 21000 #define inf 999999999 struct node { int u,v,w,next; }bian[N*40]; int head[N],cur[N],gap[N],stac[N],top,n,sink,source,y…