题意: 思路: 每回枚举去哪个山包 枚举的姿势很重要 //By SiriusRen #include <cstdio> #include <algorithm> using namespace std; int n,k,a[1005],l,r,base,cnt,ans,minn,recl,recr,recbase; int main(){ scanf("%d%d",&n,&k); for(int i=1;i<=n;i++)scanf(&qu…

要求max{F/P},先枚举下界lowf,再贪心求符合约束条件的n个最小价值和 记录F的离散值和去重可以大幅度常数优化 (本来想着用DP做的) (辣鸡POJ连auto都Complie Error) #include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<vector> #include<iterator> using name…

题目链接:http://poj.org/problem?id=3189 思路:由于题目要求最小的差值,而Range最多也才20,因此我们可以枚举上下限,多重匹配验证即可. http://paste.ubuntu.com/5943733/…

Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 40632   Accepted: 17647 Description Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the…

Fliptile Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10931   Accepted: 4029 Description Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in whic…

这题很经典啊,以前也遇到过类似的题--计蒜客 硬币翻转. 不过这题不仅要求翻转次数最少,且翻转方案的字典序也要最小. 解法:二进制枚举第一行的翻转方案,然后处理第二行,如果第二行的k列的上一列是黑色,那么第二行k列必须翻转,因为要保证当前行的上一行全为白色.在第一行确定的情况下,当前翻转一定是最优选择.一样的处理方法直到最后一行,最后检查最后一行是否有黑色,如果有说明当前方案无法成功. PS:枚举第一行是为了解题方便,枚举任何一行都可以,反正每行都能做出最优解. AC代码 #include<cs…

说好的高斯消元法呢,暴搜都能0ms 这种翻转就是枚举第一行控制变量下面行就全都确定了 代码参考挑战程序设计例题 #include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #define rep(i,j,k) for(int i=j;i<=k;i++) using namespace std; const int maxn = 1e2+11; const int d…

Treasure Hunt Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6328   Accepted: 2627 Description Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid of Key-Ops. Using state-of-the-ar…

The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 25343   Accepted: 9786   Special Judge Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a…

/* 因为15很小可以暴力枚举然后用最小生成树的prim来计算 */ #include<stdio.h> #include<string.h> #include<math.h> #define N 40 #define inf 0x3fffffff int a[N],f[N],en[N]; int ma[N][N],n,m; double dd; void prime(){ int i,j,vv[N],dis[N],sum,total; double rato; for…