qq_42479630的博客

题目链接Consider a single-elimination football tournament involving 2n teams, denoted 1, 2, …, 2n. In each round of the tournament, all teams still in the tournament are placed in a list in order of increasing index. Then, the first team in the list plays the

题目链接Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects software bugs. When Ivan gets a new program, he classifies all possible bugs into n categories. Each day he discovers exactly one bug i

题目链接题意：在一个垂直的二维平板给出n个目标的坐标，每次瞄完一个目标后实现在平板上移动的速度为1，瞄准第ii个目标后射中的概率为pi，且第i个目标只在第ti秒出现，也就是说只能在第ti秒瞄准该目标并射击，问射中目标数的期望值最大为多少？思路：当点与点之间的距离小于他们的时刻时，概率就可以相加。#include<bits/stdc++.h>using namespace st...

题目链接思路：其实也不着数位DP啦，因为它说只要开头1的可以很快算出来，我们先求出每个区间随机取一个数是开头为1的概率p【i】，然后dp【i】【j】表示前i个选了j个开头为1的概率。状态只有2个，一个是取就是p【i】，不取就是1-p【i】。这个题精度卡了我半天。。。#include<bits/stdc++.h>using namespace std;typedef long...

题目链接思路：dp【i】【j】【k】表示i场比赛赢了j场背包容量为k时的概率，转移状态也只有两个一个取一个不取。#include<bits/stdc++.h>using namespace std;typedef long long ll;const int maxn=201;double ans=0,a[maxn],dp[maxn][maxn][maxn<&lt...

题目链接#include<bits/stdc++.h>#define N 100005using namespace std;int n;double a, f1, f2;int main () { scanf ("%d", &n); for (int i = 1; i <= n; ++i) { scanf ("%lf", ...

题意：有n面镜子，作者每天都会来问魔镜，魔镜有pi/100得概率会告诉作者她很美丽，如果魔镜说美丽，那么作者会很开心，否则作者会从第一面镜子重新开始问，问最后到第n面镜子是作者是开心得期望天数。思路：设dp[i]为第i面镜子作者开心的期望天数，则dp[i]=(dp[i-1]+1)(pi/100))【开心】+(dp[i]+dp[i-1]+1)(1-pi/100)【沮丧】,移项就得dp[i]=(d...