Problem Description Mr. Panda lives in Pandaland. There are many cities in Pandaland. Each city can be treated as a point on a 2D plane. Different cities are located in different locations. There are also M bidirectional roads connecting those cities…

-->Nightmare Ⅱ 原题太复杂,直接简单的讲中文吧 Descriptions: X表示墙 .表示路 M,G表示两个人 Z表示鬼 M要去找G但是有两个鬼(Z)会阻碍他们,每一轮都是M和G先走M能走3步,G能走1步,Z每次向边上2步内变出分身.求所需最短时间. 鬼能穿墙,人不能 Sample Input 3 5 6 XXXXXX XZ..ZX XXXXXX M.G... ...... 5 6 XXXXXX XZZ..X XXXXXX M..... ..G... 10 10 ........…

Assignment Problem Description Last year a terrible earthquake attacked Sichuan province. About 300,000 PLA soldiers attended the rescue, also ALPCs. Our mission is to solve difficulty problems to optimization the assignment of troops. The assignment…

BUPT2017 wintertraining(15) #8F 题意 1到n的排列,经过几次置换(也是一个排列)回到原来的排列,就是循环了. 现在给n(<=1000),求循环周期的所有可能数. 题解 问题等价于几个正整数加起来等于n,求最小公倍数的可能数. 因为1不影响最小公倍数,所以等价于求几个正整数加起来小于等于n,最小公倍数的可能数. 最小公倍数与每个质因子在正整数里最大出现次数有关,所以枚举质因子的幂,进行dp. dp[i][j]表示前i个质数,和为j时,最小公倍数的可能数. dp[0]…

Escape  Descriptions: 一个人从(0,0)跑到(n,m),只有k点能量,一秒消耗一点,在图中有k个炮塔,给出炮塔的射击方向c,射击间隔t,子弹速度v,坐标x,y问这个人能不能安全到达终点要求: 1.人不能到达炮塔所在的坐标2.炮塔会挡住子弹3.途中遇到子弹是安全的,但是人如果停在这个坐标,而子弹也刚好到这个坐标,人就被射死4.人可以选择停止不动 Input 对于每个测试用例,第一行有四个整数,m.n.k和d (2<=m, n<=100, 0<=k<=100, m…

Vacation 题意 有n+1辆车,属性有长度l,距离终点的距离s,速度v问你最末尾的车到达终点的时间 Sample Input 1 2 2 7 1 2 1 2 1 2 2 10 7 1 6 2 1 Sample Output 3.5000000000 5.0000000000 注意哈  Sample Input改一下格式 1 2 2 71 2 1 2 1 2 2 10 7 1 6 2 1 这样看的就舒服写了  我一开始做题,硬是看了数据好长时间,感觉这是故意这样写的,好坑啊 题目链接 htt…

题目给定n和d,都是10的9次方以内,求1到n里面有几个数最大因数是d?1000000组数据.解:求出d的满足p[i]*d<n的最小质因数是第几个质数.即为答案. #include<cstdio> #define N 100002 int t,n,d,pr[N],p[N],num; int main(){ for(int i=2;i<N;i++)if(!pr[i]){ for(int j=i+i;j<N;j+=i) pr[j]=1; p[++num]=i; } scanf(&…

题意是给你抢劫每个银行可获得的钱m和被抓的概率p,求被抓的概率小于P,最多能抢多少钱.01背包问题,体积是m,价值是p.被抓的概率不是简单相加,而应该是1−Π(1−p[i])DP:dp[i]表示抢到i元且不被抓的概率.初始条件:dp[0]=1:转移方程:dp[j]=max{dp[j],dp[j−m[i]]∗(1−p[i])}(j>=m[i]) #include<cstdio> #include<cstring> #include<algorithm> #defin…

Problem Description Mr.Panda is about to engage in his favourite activity doing laundry! He's brought L indistinguishable loads of laundry to his local laundromat, which has N washing machines and M dryers.The ith washing machine takes Wi minutes to…

[UOJ#311][UNR #2]积劳成疾(动态规划) UOJ Solution 考虑最大值分治解决问题.每次枚举最大值所在的位置,强制不能跨过最大值,左右此时不会影响,可以分开考虑. 那么设\(f[i][j]\)表示长度为\(i\),且最大值不超过\(j\)的所有方案之和. 因为最大值有多个,所以我们钦定每次选择最靠右的那个,所以转移就是: \[f[i][j]=f[i][j-1]+\sum_{k=1}^if[k-1][j]*f[i-k][j-1]*w[j]^{c}\] 即钦定为最靠右的那个最大…