[题目链接] http://poj.org/problem?id=2100 [题目大意] 给出一个数,求将其拆分为几个连续的平方和的方案数 [题解] 对平方数列尺取即可. [代码] #include <cstdio> using namespace std; typedef long long LL; const int N=10000010; LL n,ansl[N],ansr[N]; int main(){ while(~scanf("%lld",&n)){ L…

墓地 题目大意,给定一个整数,要你找出他的平方和组合 太简单了....不过一开始我储存平方和想降低时间,后来发现会超内存,直接用时间换空间了,游标卡尺法 #include <iostream> #include <functional> #include <algorithm> #define MAX_N 10000001 using namespace std; typedef long long LL_INT; ];//只储存开始和结尾 void Inivilize…

Graveyard Design Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 4443   Accepted: 946 Case Time Limit: 2000MS Description King George has recently decided that he would like to have a new design for the royal graveyard. The graveyard mu…

直接枚举就行了 #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #include<vector> #define ll __int64 #define pi acos(-1.0) #define SIZE 1024 using namespace std;…

Description King George has recently decided that he would like to have a new design for the royal graveyard. The graveyard must consist of several sections, each of which must be a square of graves. All sections must have different number of graves.…

Graveyard Design Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 6107   Accepted: 1444 Case Time Limit: 2000MS Description King George has recently decided that he would like to have a new design for the royal graveyard. The graveyard m…

尺取法:顾名思义就是像尺子一样一段一段去取,保存每次的选取区间的左右端点.然后一直推进 解决问题的思路: 先移动右端点 ,右端点推进的时候一般是加 然后推进左端点,左端点一般是减 poj 2566 题意:从数列中找出连续序列,使得和的绝对值与目标数之差最小 思路: 在原来的数列开头添加一个0 每次找到的区间为 [min(i,j)+1,max(i,j)] 应用尺取法的代码: while (r <=n) { int sub = pre[r].sum - pre[l].sum; while (abs(…

POJ 1852 Ants POJ 2386 Lake Counting POJ 1979 Red and Black AOJ 0118 Property Distribution AOJ 0333 Ball POJ 3009 Curling 2.0 AOJ 0558 Cheese POJ 3669 Meteor Shower AOJ 0121 Seven Puzzle POJ 2718 Smallest Difference POJ 3187 Backward Digit Sums POJ 3…

http://poj.org/problem?id=3181 Description Farmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are selling variously for $1, $2, and $3. Farmer John has exactly $5 to…

poj 题目分类 按照ac的代码长度分类(主要参考最短代码和自己写的代码) 短代码:0.01K--0.50K:中短代码:0.51K--1.00K:中等代码量:1.01K--2.00K:长代码:2.01K以上. 短:1147.1163.1922.2211.2215.2229.2232.2234.2242.2245.2262.2301.2309.2313.2334.2346.2348.2350.2352.2381.2405.2406: 中短:1014.1281.1618.1928.1961.2054…