Problem Description Mickey is interested in probability recently. One day , he played a game which is about probability with mini.First mickey gives a letter and a word to mini.Then mini calculate the probability that the letter appears in the word.F…
http://acm.hdu.edu.cn/showproblem.php?pid=2131 Problem Description Mickey is interested in probability recently. One day , he played a game which is about probability with mini.First mickey gives a letter and a word to mini.Then mini calculate the pr…
Probability 链接:http://acm.hdu.edu.cn/showproblem.php?pid=2131 Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8731 Accepted Submission(s): 4228 Problem Description Mickey is interested in pro…
FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784 Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…
Nasty Hacks Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049 Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…
Box of Bricks Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994 Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…
[BZOJ2318]Spoj4060 game with probability Problem Description Alice和Bob在玩一个游戏.有n个石子在这里,Alice和Bob轮流投掷硬币,如果正面朝上,则从n个石子中取出一个石子,否则不做任何事.取到最后一颗石子的人胜利.Alice在投掷硬币时有p的概率投掷出他想投的一面,同样,Bob有q的概率投掷出他相投的一面. 现在Alice先手投掷硬币,假设他们都想赢得游戏,问你Alice胜利的概率为多少. Input 第一行一个正整数t,…
Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color…
卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…
