题目链接 卡了一下精度和内存. #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <iostream> #include <algorithm> using namespace std; #define eps 1e-8 #define N 1000001 #define LL __int64 ] = {-,,,-,,,-,…

以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight Moves1101 Gamblers1204 Additive equations 1221 Risk1230 Legendary Pokemon1249 Pushing Boxes 1364 Machine Schedule1368 BOAT1406 Jungle Roads1411 Annive…

/* poj 1654 Area 多边形面积 题目意思很简单,但是1000000的point开不了 */ #include<stdio.h> #include<math.h> #include<string.h> const int N=1000000+10; const double eps=1e-8; struct point { double x,y; point(){} point(double a,double b):x(a),y(b){} }; int le…

链接:http://poj.org/problem?id=1654 Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14952   Accepted: 4189 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orth…

Area Time Limit: 1000MS Memory Limit: 10000K Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From this vertex, you may go step by step to the fol…

题意:找到一段数字里最大值和最小值的差 水题 #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<queue> using namespace std; ; const int INF=0x3f3f3f3f; int n,m,t; ; ],dpMIN[MAXN][]; int mm[MAXN…

一.Description As part of an arithmetic competency program, your students will be given randomly generated lists of from 2 to 15 unique positive integers and asked to determine how many items in each list are twice some other item in the same list. Yo…

水题直接码... /********************* Template ************************/ #include <set> #include <map> #include <list> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16894   Accepted: 4698 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From thi…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17456   Accepted: 4847 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From thi…

一个简单的用叉积求任意多边形面积的题,并不难,但我却错了很多次,double的数据应该是要转化为long long,我转成了int...这里为了节省内存尽量不开数组,直接计算,我MLE了一发...,最后看了下别人的才过,我的代码就不发了,免得误导,不得不说几何真是... 还有就是这个大神的代码,貌似G++,过不了,C++AC #include <iostream> #include <algorithm> #include <cstdio> #include <c…

Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From this vertex, you may go step by step to the following vertexes of the polygon until back to…

Going Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15944   Accepted: 8167 Description On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertical…

题目 http://poj.org/problem?id=1837 题意 单组数据,有一根杠杆,有R个钩子,其位置hi为整数且属于[-15,15],有C个重物,其质量wi为整数且属于[1,25],重物与重物之间,钩子与钩子之间彼此不同.忽略杠杆及重心的影响,有多少种方式使得全部重物都挂上钩子(某些钩子可能挂若干个重物)后杠杆平衡? 思路 由于状态比较小,即使n的五次方也足以承受,而且任意时刻杠杆的状态在[-15 * 25 * 20, 15 * 25 * 20]之间,所以可以直接穷举状态. 感想…

#include<stdio.h> #include<string.h> #include<iostream> #include<math.h> using namespace std; ]={,,,,,,,-,-,-}; ]={,-,,,-,,,-,,}; ]; __int64 area,x,y,px,py; int main() { int sum,t,tmp,i; cin>>tmp; while(tmp--) { scanf("%…

大概题意就是求\(1 \le i,j \le n\)的\(gcd(i,j) = 1\)的个数+2(对于0的特判) 正解应该是欧拉函数或者高逼格的莫比乌斯反演 但数据实在太水直接打表算了 /*H E A D*/ bool GCD[1002][1002]; inline int gcd(int a,int b){return b?gcd(b,a%b):a;} int main(){ rep(i,1,1000) rep(j,1,1000) GCD[i][j]=bool(gcd(i,j)==1); in…

题意: 找出这些串中最长的公共子串(长度≥3),如果长度相同输出字典序最小的那个. 分析: 用库函数strstr直接查找就好了,用KMP反而是杀鸡用牛刀. #include <cstdio> #include <cstring> ][], sub[]; ], l; int cmp(int p1, int p2) { ; i < l; ++i) ][p1 + l] > a[][p2 + l]) return p2; return p1; } int main(void)…

[题意简述]:给出我们钉子个数与半径,让我们求出缠绕在钉子上的绳子有多长. [分析]:从题目中我们能够看出,绳子长度的和等于每两个钉子的距离的和加上接触在钉子上的绳子的长度,不难发现这部分长度事实上就等于钉子的周长. 见代码: #include<iostream> #include<cmath> using namespace std; #define Pi 3.1415//这个精度要尽量高! 也能够用4.0*atan(1.0) double s(double x1,double…

题意:从原点出发,沿着8个方向走,每次走1个点格或者根号2个点格的距离,最终回到原点,求围住的多边形面积. 分析:直接记录所经过的点,然后计算多边形面积.注意,不用先保存所有的点,然后计算面积,边走变算,不然会超内存.最多有1000000个点. 注意:精度问题,使用long long /__int64,直接使用double不准确.方向的处理使用数组. // Time 94ms; Memory 1036K #include<iostream> #include<cstring> #d…

Counterfeit Dollar Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 35774   Accepted: 11390 Description Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit ev…

题目大意就是有很多牛.告诉你每只牛的高度.然后有很多个询问.输出该区间内的最大身高差.也就是用RMQ求最大值最小值.貌似还可以用线段树.然而,我还不会线段树.....T_T 可能是太多组数据了.cin和cout会TLE.换成scanf和printf就顺当的AC了....啦啦啦. RMQ还是只会用模板..T_T 附代码:#include<stdio.h>#include<string.h>#include<iostream>#include<math.h>#d…

Ants Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 10722   Accepted: 4752 Description An army of ants walk on a horizontal pole of length l cm, each with a constant speed of 1 cm/s. When a walking ant reaches an end of the pole, it imm…

Rounders Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7697   Accepted: 4984 Description For a given number, if greater than ten, round it to the nearest ten, then (if that result is greater than 100) take the result and round it to th…

题意: 从一个点出发,8个方向,给出每一步的方向,求出走过的路径形成的多边形的面积. 思路: 先普及一下向量叉乘.. (摘自度娘) 也就是x1y2-x2y1. 那这不就好说了嘛. 一个经过原点的闭合多边形的面积可以分割成多个三角形. 每个三角形的面积就是相邻向量叉乘的一半. 大功告成~ 注意精度问题... // by SiriusRen #include <cstdio> #include <cstring> using namespace std; char a[1000005]…

读懂题意就简单了 #include<stdio.h> #define inf 999999999 #define N 310 int f[N]; int map[N][N]; int main() { int n,m,i,j,k,a,b,max,ans; while(scanf("%d%d",&n,&m)!=EOF) { for(i=1;i<=n;i++) for(j=1;j<=n;j++) map[i][j]=inf; while(m--) {…

枚举点 每次都搜一遍 //By SiriusRen #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define N 20005 int ans,k,n,m,first[N],next[N],v[N],tot,xx,yy,mark[N],vis[1005]; void add(int x,int y){ v[tot]=y,next[tot]=first[x],…

#include<stdio.h> #include<algorithm> #include <cstring> using namespace std; typedef long long ll; const int MAXN = 1000008; char s[MAXN]; int dx[] = {-1, -1, -1, 0, 0, 0, 1, 1, 1}; int dy[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1}; int main()…

I Think I Need a Houseboat Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 85149   Accepted: 36857 Description Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the land,…

POJ 1488 题目大意:给定一篇文章,将它的左引号转成 ``(1的左边),右引号转成 ''(两个 ' ) 解题思路:水题,设置一个bool变量标记是左引号还是右引号即可 /* POJ 1488 Tex Quotes --- 水题 */ #include <cstdio> #include <cstring> int main() { #ifdef _LOCAL freopen("D:\\input.txt", "r", stdin); #…

POJ 3176 Cow Bowling 链接: http://poj.org/problem?id=3176 这道题可以算是dp入门吧.可以用一个二维数组从下向上来搜索从而得到最大值. 优化之后可以直接用一维数组来存.(PS 用一维的时候要好好想想具体应该怎么存,还是有技巧的) #include<iostream> #include<cstring> #include<cmath> #include<cstdio> using namespace std;…